Algebra

We can add, subtract and multiply complex numbers: If \(z_{1}=x_{1}+iy_{1}\) and \(z_{2}=x_{2}+iy_{2}\) then \[z_{1}\pm z_{2}:=(x_{1}\pm x_{2})+i(y_{1}\pm y_{2}),\] \[z_{1}z_{2}:=(x_{1}x_{2}-y_{1}y_{2})+i(x_{1}y_{2}+x_{2}y_{1}).\] Notice that addition simply corresponds to adding the individual components. In general we denote by \(\mathrm{Re}(z)=x\) the real part of \(z\), and by \(\mathrm{Im}(z)=y\) the imaginary part of \(z\). By the definition of multiplication we have \(i^{2}=-1\), and using this we see that multiplication corresponds to ‘multiplying out the brackets’: \((x_{1}+iy_{1})(x_{2}+iy_{2})=(x_{1}x_{2}+i^{2}y_{1}y_{2})+i(x_{1}y_{2}+x_{2}y_{1}).\)

We can also divide complex numbers. For \(z_{2}\neq0\) (here we use the shorthand \(0=0+0i\)) we have \[\frac{z_{1}}{z_{2}}=\frac{x_{1}+iy_{1}}{x_{2}+iy_{2}}=\frac{(x_{1}+iy_{1})(x_{2}-iy_{2})}{(x_{2}+iy_{2})(x_{2}-iy_{2})}=\frac{x_{1}x_{2}+y_{1}y_{2}}{x_{2}^{2}+y_{2}^{2}}+i\frac{x_{2}y_{1}-x_{1}y_{2}}{x_{2}^{2}+y_{2}^{2}}\quad\in\mathbb{C}.\] The quantity we used to make the denominator real is important. In general, for \(z=x+iy\) we call \(\bar{z}:=x-iy\) the complex conjugate of \(z\).

We immediately have a multiplicative inverse \[z^{-1}:=\frac{1}{z}=\frac{x}{x^{2}+y^{2}}-i\frac{y}{x^{2}+y^{2}}.\] WARNING: While most of the nice properties of \(\mathbb{R}\) hold in \(\mathbb{C}\), we do not have notions of \(\leq\) \(<\), \(\geq\) or \(>\); the set \(\mathbb{C}\) is not ‘ordered’ and expressions like \(z_{1}<z_{2}\) have no meaning.

How do we visualise the complex numbers?

There are various ‘models’ for the complex numbers. The most commonly used/most intuitive is to think of \(\mathbb{C}\) as a copy of \(\mathbb{R}^{2}\) equipped with a map \[\mathbb{R}^{2}\times\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}:\left((x_{1},y_{1}),(x_{2},y_{2})\right)\mapsto(x_{1}x_{2}-y_{1}y_{2},x_{1}y_{2}+x_{2}y_{1}).\] So, a copy of \(\mathbb{R}^{2}\) with a way of multiplying² (and dividing) vectors! Indeed there is an obvious bijection \(f:\mathbb{R}^{2}\rightarrow\mathbb{C}\) given by \(f((x,y))=x+iy\). As a result, we often draw complex numbers on the usual \((x,y)\)-plane: such a picture is called an Argand diagram.

On \(\mathbb{R}^{2}\) there is a natural notion of size, and we use it in \(\mathbb{C}\): we call the quantity \(|z|:=\sqrt{x^{2}+y^{2}}\) the modulus or absolute value of \(z\) \((=x+iy)\).

Polar form/coordinates of a complex number

Now we have a notion of distance, as in \(\mathbb{R}^{2}\) we can implement a change of variables \(z(x,y)\rightarrow z(r,\theta)\). Let \(r=|z|\) and let \(\theta\) denote the anticlockwise angle measured from the real axis. (Angles measured clockwise will be considered negative.) We call \(\theta\) the argument of \(z\) (for \(z\neq0\)) and write \(\arg(z)=\theta\). We then have the following polar coordinates for \(z\): \[z=r(\cos\theta+i\sin\theta),\] which we write in shorthand as \(z=re^{i\theta}\).

For example, \(i=e^{i\pi/2}\) and \(1+i=\sqrt{2}e^{i\pi/4}\). We have \(|-1/\sqrt{2}-i\sqrt{3}/\sqrt{2}|=\sqrt{2}\), so \[-1/\sqrt{2}-i\sqrt{3}/\sqrt{2}=\sqrt{2}(-1/2-i\sqrt{3}/2)=\sqrt{2}e^{-i2\pi/3}.\] Note that \(\arg(z)\) is only defined up to multiples of \(2\pi\); for example \(i=e^{i\pi/2}=e^{i5\pi/2}=e^{-i3\pi/2}\). Strictly speaking \(\arg(i)=\pi/2+2\pi k\), for any \(k\in\mathbb{Z}\) (and so \(\arg\) is a one-to-many function!). As a result, we need to be careful; we choose a fixed interval in which to express the argument: the principal value of \(\arg(z)\) is the value in the interval \((-\pi,\pi]\) and will be denoted \(\mathrm{Arg}(z)\). So \(\mathrm{Arg}(i)=\pi/2\) and \(\mathrm{Arg}(-1)=\pi\) for example.

It is nice to have a geometric picture of what the algebraic operations on complex numbers mean.

The modulus also has the following important properties.

1. (Triangle inequality) \(|z_{1}+z_{2}|\leq|z_{1}|+|z_{2}|\)

2. \(|z|\geq0\text{ and }|z|=0\iff z=0\)

3. \(\max(|\mathrm{Re}(z)|,|\mathrm{Im}(z)|)\leq|z|\leq|\mathrm{Re}(z)|+|\mathrm{Im}(z)|\)

Simple complex functions and geometry (examples)

• We can also use functions to define regions in the complex plane. Consider the set of points \(z\) which satisfy the inequality \(|z-i|<|z+i|\). This is precisely the points in \(\mathbb{C}\) whose distance to \(i\) is strictly smaller than their distance to \(-i\). Thus, the inequality represents the upper half plane \(\mathbb{H}:=\{z\in\mathbb{C}:\mathrm{Im}(z)>0\}\).

• Note that the equation \(|z-i|=1\) represents a circle centred at \(i\) of radius \(1\).

Finding values of log and complex powers (examples)

\((a)\) Using the principal branch of \(\log\), we find \(\log(1-i)\) and \((1-i)^{1/2}\). We have \(|1-i|=\sqrt{2}\) and \(\mathrm{Arg}(1-i)=-\pi/4\). Thus, \(1-i=\sqrt{2}e^{-i\pi/4}\). Therefore, \(\mathrm{Log}(1-i)=\log|1-i|+i\mathrm{Arg}(1-i)=\log\sqrt{2}-i\pi/4\), and \[(1-i)^{1/2}=\exp\left(\frac{1}{2}\,\mathrm{Log}(1-i)\right)=\exp\left(\frac{1}{2}\,\left(\log\sqrt{2}-i\frac{\pi}{4}\right)\right)=\exp\left(\log\sqrt[4]{2}-i\frac{\pi}{8}\right)=\sqrt[4]{2}e^{-i\frac{\pi}{8}}.\] \((b)\) Using the principal branch and the previous example \[(1-i)^{i}=\exp(i\mathrm{Log}(1-i))=\exp\left(\frac{\pi}{4}+i\log\sqrt{2}\right)=e^{\pi/4}e^{i\log\sqrt{2}}.\] \((c)\) Again, using the principal branch \[2^{1/2}=\exp\left(\frac{1}{2}\log2\right)=\exp(\log\sqrt{2})=\sqrt{2}.\] What about the other root? It comes from using a different branch: if we let \(\log\) be the branch of logarithm corresponding to \(\arg(z)\in(\pi,3\pi]\) then \(\log(z)=\log|z|+i(\mathrm{Arg}(z)+2\pi)\), so we have \(\log(2)=\log2+i2\pi\) and \[2^{1/2}=\exp\left(\frac{1}{2}(\log2+i2\pi)\right)=\exp(\log\sqrt{2}+i\pi)=\sqrt{2}e^{i\pi}=-\sqrt{2}.\]

Visualizing complex functions

The ‘graph’ of a complex-valued function \(f:\mathbb{C}\to\mathbb{C}\) is 4-dimensional, so difficult to visualise - we certainly can’t draw it. We can employ other techniques to get a grasp on complex functions:

• We can graph the real-valued function \(|f|:\mathbb{C}\to\mathbb{R}\). For example, consider the complex function \(\cos z\). When \(z=x\) is purely real, we have that \(|f(z)|=|\cos(x)|\) is obviously just the modulus of the real cosine function. But for \(z=iy\) purely complex we have \(|f(z)|=|\cos(iy)|=|\cosh y|=\cosh y.\) So in the imaginary direction \(f\) simply looks like cosh!

• It is often useful to visualise complex functions by considering how they map regions of the complex plane. Consider the image of the ‘right half-plane’ \(\mathbb{H}_{R}:=\{z\in\mathbb{C}:\mathrm{Re}(z)>0\}\) under the map \(f(z)=z^{2}\). Note that \(\mathbb{H}_{R}=\{z\in\mathbb{C}:-\pi/2<\mathrm{Arg}(z)<\pi/2\}\). If \(z=re^{i\theta}\in\mathbb{H}_{R}\) then \(z^{2}=zz=r^{2}e^{i2\theta}\) has argument \(2\theta\in(-\pi,\pi)\). Thus, \(f\) maps \(\mathbb{H}_{R}\) to \(\mathbb{C}\setminus\mathbb{R}_{\leq0}\), where \(\mathbb{R}_{\leq0}\) denotes the negative real axis (including \(0\)). The map is onto, since for every \(w=se^{i\phi}\in\mathbb{C}\setminus\mathbb{R}_{\leq0}\) (so \(\phi\in(-\pi,\pi)\)) we can find \(z\in\mathbb{H}_{R}\) such that \(f(z)=w\); namely we can choose \(z=\sqrt{s}e^{i\phi/2}\in\mathbb{H}_{R}\).

  Similarly, the left half plane \(\mathbb{H}_{L}=\{z\in\mathbb{C}:\mathrm{Re}(z)<0\}\) is mapped to \(\mathbb{C}\setminus\mathbb{R}_{\leq0}\). Moreover, \(f\) maps both the strictly positive imaginary axis \(i\mathbb{R}_{>0}\) and the strictly negative imaginary axis \(i\mathbb{R}_{<0}\) (given by \(i\mathbb{R}_{>0}=\{iy\in\mathbb{C}:y>0\}\) and \(i\mathbb{R}_{<0}=\{iy\in\mathbb{C}:y<0\}\) respectively) to the strictly negative real axis \(\mathbb{R}_{<0}\).

  Adding the observation that \(f(0)=0\), we have that \(f(z)=z^{2}\) in essence maps \(\mathbb{C}\) to two copies of itself (except for the origin, which is only attained once in the image - remember this different behaviour at the origin later!)

• Branches of log. Let \(\log\) be the branch of logarithm corresponding to \(\arg(z)\in(\theta_{1},\theta_{2}]\). Then \(\log\) maps \(\mathbb{C}\setminus R_{\theta_{1}}\) to the infinite horizontal strip \[\{z\in\mathbb{C}\::\:\theta_{1}<\mathrm{Im}(z)\leq\theta_{2}\}.\] Infinite rays emanating from \(0\) map to horizontal lines, and circles centred at zero, minus their interection with \(R_{\theta_{1}}\), map to vertical line segments.

The Riemann sphere

Consider the unit sphere \(S^{2}:=\{(x,y,s)\in\mathbb{R}^{3}:x^{2}+y^{2}+s^{2}=1\}\) in \(\mathbb{R}^{3}\) and consider a copy of \(\mathbb{C}\) embedded in \(\mathbb{R}^{3}\) by identifying \(\mathbb{C}=\mathbb{R}^{2}\) with the \((x,y)\)-plane. Explicitly, a point \(x+iy\in\mathbb{C}\) corresponds to the point \((x,y,0)\in\mathbb{R}^{3}\).

Let \(N=(0,0,1)\in S^{2}\) denote the ‘north pole’. For any point \(v\in S^{2}\setminus\{N\}\), there is a unique straight line \(L_{N,v}\) passing through \(N\) and \(v\). Since \(v\neq N\), this line is not parallel to the \((x,y)\)-plane. Hence it intersects the \((x,y)\)-plane in a unique point \((x,y,0)\). This corresponds to the point \(x+iy\in\mathbb{C}\). We have defined a map \(P:S^{2}\setminus\{N\}\to\mathbb{C}\) by \(P(v)=x+iy\) in the notation of the preceding discussion. The map \(P\) is called the stereographic projection (from the north pole).

What is the formula for stereographic projection? Let \((x,y,s)\in S^{2}\setminus\{N\}\). Note that \(s\neq1\). The equation of the line passing through the point \((x,y,s)\in S^{2}\) and the North Pole \(N=(0,0,1)\in S^{2}\) is given by \[\gamma(t)\:=\:N+\left(\begin{pmatrix}x\\ y\\ s \end{pmatrix}-N\right)\,t\:=\:\begin{pmatrix}0\\ 0\\ 1 \end{pmatrix}+\begin{pmatrix}x\\ y\\ s-1 \end{pmatrix}\,t,\quad\quad\quad(t\in\mathbb{R}).\] This clearly intersects the plane when \(t=\frac{1}{1-s}\). Thus \[P:(x,y,s)\mapsto\frac{x}{1-s}+\frac{iy}{1-s}.\]

It is possible to find an inverse to \(P\): given any point \(z\in\mathbb{C}\), draw the straight line passing through \(N\) and \(z\) (thinking of \(\mathbb{C}\) as the \((x,y)\)-plane as before). This straight line passes through \(S^{2}\) in exactly one point. Hence \(P\) is a bijection that identifies \(S^{2}\setminus\{N\}\) with \(\mathbb{C}\).

Now we have identified \(\mathbb{C}\) with \(S^{2}\setminus\{N\}\) via the map \(P\), it gives us a natural way to view the added point \(\infty\) of \(\hat{\mathbb{C}}\). It should correspond to adding back in the north pole to \(S^{2}\setminus\{N\}\). In other words, we should think of \(\hat{\mathbb{C}}\) simply as the entire sphere \(S^{2}\)!

In fact, one can show that we have the following correspondences/mappings: \[\begin{align} \underline{\text{In } S^{2}} & & \underline{\text{In } \hat{\mathbb{C}}}\label{eq:s2-table1}\\ N & \longleftrightarrow & \infty\nonumber \\ S & \longleftrightarrow & 0\nonumber \\ \text{Equator } & \longleftrightarrow & \text{Unit circle }\{z\in\mathbb{C}:|z|=1\}\nonumber \\ \text{(open) Southern hemisphere} & \longleftrightarrow & \mathbb{D}:=\{z\in\mathbb{C}:|z|<1\}\nonumber \\ \text{(open) Northern hemisphere} & \longleftrightarrow & \hat{\mathbb{C}}\setminus\overline{B}_{1}(0)=\hat{\mathbb{C}}\setminus\{z\in\mathbb{C}:|z|\leq1\}\nonumber \\ & \text{Algebraically:}\nonumber \\ (x,y,s) & \longleftrightarrow & \frac{x+iy}{1-s}\quad\:\:\text{(Stereo. Proj.)}\nonumber \\ \frac{1}{|z|^{2}+1}\left(2\mathrm{Re}(z),2\mathrm{Im}(z),|z|^{2}-1\right) & \longleftrightarrow & z\quad\quad\quad\quad\text{(Inverse Stereo. Proj.)}\nonumber\end{align}\]

Examples of Metrics

1. The metric induced by the modulus function \(|\,.\,|\) on \(\mathbb{R}\) or \(\mathbb{C}\). We can define a distance function \(d\) on \(\mathbb{R}\times\mathbb{R}\) or \(\mathbb{C}\times\mathbb{C}\) by the formula \(d(x,y)=|x-y|\). This metric satisfies (D1)-(D3) by the properties \(1.\) and \(2.\) of the modulus we gave just after Corollary [cor:norm-mult].

2. The Euclidean norm on \(\mathbb{R}^n\) or \(\mathbb{C}^n\) For vectors \(\boldsymbol{x}\) in \(\mathbb{R}^{n}\) (or \(\mathbb{C}^{n}\)), the function \[d(\boldsymbol{x},\boldsymbol{y}):=\|\boldsymbol{x}-\boldsymbol{y}\|_{2}=\sqrt{\sum_{i=1}^{n}|x_{i}-y_{i}|^{2}}\] is a metric. It is easy to check properties \((D1)-(D3)\). Note that this norm comes from an inner-product. For example, for \(n=2\), the real Euclidean norm on \(\mathbb{R}^{2}\) comes from the usual dot product \(\|\boldsymbol{x}\|_{2}=\sqrt{\boldsymbol{x}\cdot\boldsymbol{x}}\) and the complex Euclidean norm on \(\mathbb{C}^{2}\) comes from the inner product \(\left<z,w\right>=z_{1}\bar{w}_{1}+z_{2}\bar{w}_{2}\); that is \(\|z\|_{2}=\sqrt{\left<z,z\right>}\).

3. • (a) Metrics induced from inner products in vector spaces

     More generally, given any finite dimensional real vector space \(V\) with a (positive definite) inner product \(\left<\,.\,\right>\), then \[d(v,w):\,=\,\|v-w\|\,=\,\sqrt{\left<v-w,v-w\right>}\quad\quad(v,w\in V)\] is a metric. Properties (D1) and (D2) are obvious, property (D3) follows from Cauchy-Schwarz: \(|\left<v,w\right>|\leq\|v\|\cdot\|w\|\) - see sheet 2.

   • (b) Metrics induced from norms in vector spaces

     Even more generally, so long as a vector space has a ‘nice’ notion of the ‘size’ of each vector, we can define a metric in the obvious way. Such a notion is in generality referred to as a norm:

     [Norms and normed vector spaces] Given any real or complex vector space \(V\), a function \(\|\,.\,\|:V\to\mathbb{R}_{\geq0}\) is a norm if it satisfies (for \(v,w\in V\))

     • (N1) \(\|v\|\geq0\) and \(\|v\|=0\quad\Longleftrightarrow\quad v=0;\)

     • (N2) \(\|\lambda v\|=|\lambda|\cdot\|v\|\quad\) for \(\lambda\in\mathbb{R}\) or \(\mathbb{C}\);

     • (N3) \(\|v+w\|\leq\|v\|+\|w\|\) (the triangle inequality).

     Note that (N3) implies \(\|v-w\|\geq\left|\,\|v\|-\|w\|\,\right|\) (the reverse triangle inequality). A vector space equipped with a norm is called a normed vector space. The metric given by \(d(v,w):=\|v-w\|\) then always defines a metric (it is easy to check properties (D1)-(D3)). In particular, since the modulus function on \(\mathbb{C}\) is a norm, the metric we get from the modulus function comes from a norm.

4. \(\ell_p\)-norm on \(\mathbb{R}^n\) or \(\mathbb{C}^n \, (p \geq 1)\)

   The above suggests that a vector space could be home to many different norms (so many different metrics). But, not all norms arise from an inner product as in \((3a)\); for example, for vectors \(\boldsymbol{x}\) in \(\mathbb{R}^{n}\) or \(\mathbb{C}^{n}\), the function \[\|\boldsymbol{x}\|_{p}:=\sqrt[p]{\sum_{i=1}^{n}|x_{i}|^{p}}\] defines a norm for every \(p\geq1\), called the \(\boldsymbol{\ell_{p}}\)-norm. But, for \(p\ne2\) (the Euclidean norm) it does not arise from an inner product. When \(p=1\), the \(\ell_{1}\)-norm is simply given by the sum of the size of the components \(\|\boldsymbol{x}\|_{1}=\sum_{i=1}^{n}|x_{i}|\) and is sometimes referred to as the Taxicab norm.

5. \(\ell_\infty\)-norm on \(\mathbb{R}^n\) (or \(\mathbb{C}^n\))

   The function \[\|\boldsymbol{x}\|_{\infty}:=\max_{i=1,\ldots,n}{|x_{i}|}\] also defines a norm, called the \(\boldsymbol{\ell_{\infty}}\)-norm (or the sup-norm), thus it also defines a metric. It is in some sense the ‘limiting notion’ of the \(\ell_{p}\) norms.

6. Riemannian (chordal) metric on \(\hat{\mathbb{C}}\)

   Let \(f:\hat{\mathbb{C}}\to S^{2}\) be the (inverse of the) stereographic projection. Then the function \[d(z,w):=\|f(z)-f(w)\|_{2}\quad\quad(z,w\in\hat{\mathbb{C}}),\] where \(\|\,.\,\|_{2}\) is Euclidean norm in \(\mathbb{R}^{3}\) (so, the \(\ell_{2}\)-norm), is a metric on \(\hat{\mathbb{C}}\). It is called the Riemannian metric (or chordal metric). Note that with respect to this metric, the distance from \(0\) to \(i\) is the same as the distance from \(i\) to \(\infty\), for example!

7. Discrete metric

   Let \(X\) be a non-empty finite set. Then (for \(x,y\in X\)) the function \[d(x,y):=\begin{cases} 0\ & \text{ if }\quad x=y,\\ 1\ & \text{ if }\quad x\neq y, \end{cases}\] defines a metric, called the discrete metric. It is easy to check (D1)-(D3). In this case, \((X,d)\) is called a discrete metric space.

8. Function spaces

   There are many of these, such as the space \(X=C([a,b])\) of continuous functions on an interval \([a,b]\). The function \[\|f\|:=\max_{x\in[a,b]}|f(x)|\] defines a norm, and thus a metric (see Analysis III for more examples).

9. Subspace metric

   Any non-zero subset \(Y\subset X\) of a metric space \(X\) is itself a metric space with respect to the same metric (this is easy to check). The metric restricted to the set \(Y\) is then called the subspace metric [There is actually much more than meets the eye with this metric - see sheets 2 and 3.]

Visualizing balls (examples)

1. Let \(X=\mathbb{C}\) and \(d(z,w)=|z-w|\), then \(B_{1}(0)=\mathbb{D}=\{z\::\:|z|<1\}\) as before. More generally \(B_{r}(z_{0})\) is the usual ball of radius \(r\) around \(z_{0}\), not including its boundary circle. \(\bar{B}_{r}(z_{0})\) is the ball of radius \(r\) around \(z_{0}\), including its boundary circle. This is the most important example from the point of view of Complex Analysis.

2. Let us consider the unit ball \(B_{1}(\mathbf{0})\) in \(\mathbb{R}^{2}\) with respect to the \(\ell_{p}\)-norms, for \(p=1,2\) and \(\infty\).

   For \(p=2\) the unit ball \(B_{1}(\mathbf{0})\) is the usual Euclidean ball - so the inside of the unit circle centred at the origin. For \(p=\infty\), the equation \(\max\{|x|,|y|\}<1\) (for \((x,y)\in\mathbb{R}^{2}\)) clearly defines the interior of a square with vertices \((1,1),(-1,1),(1,-1)\) and \((-1,-1)\).

   For the \(\ell_{1}\)-norm a little care is needed. We are interested in the points \((x,y)\in\mathbb{R}^{2}\) with \(|x|+|y|<1\). In the 1st quadrant this means \(y<1-x\), in the 2nd it means \(y<1+x\), in the 3rd we have \(y>-1-x\) and in the 4th its \(y>x-1\). Thus, the unit ball is the interior of a diamond with vertices \((1,0),(0,1),(-1,0)\) and \((0,-1)\).

• A subset \(U\subseteq X\) is open (in \(\boldsymbol{X}\)) if for every point \(x\in U\) there exists \(\epsilon>0\) such that \(B_{\epsilon}(x)\subset U\).

• A subset \(U\subseteq X\) is closed (in \(\boldsymbol{X}\)) if its complement \(X\setminus U\) is open.

Open sets (examples/warnings)

1. All of the previously encountered subsets \(\mathbb{H},\mathbb{D},\mathbb{C}^{\ast}\) and \(\mathbb{C}\setminus\mathbb{R}_{\leq0}\) of the complex plane are open (see sheet 2). The 1st quadrant \(\Omega_{1}:=\{z\in\mathbb{C}:\mathrm{Re}(z)>~0,\mathrm{Im}(z)>0\}\) is open. To see this, for \(z\in\Omega_{1}\) consider the ball \(B_{r}(z)\) where \(r=\min(\mathrm{Re}(z),\mathrm{Im}(z))/2\).

2. Let \(X\) be a discrete metric space, so \(d(x,y):=\begin{cases} 0\ & \text{ if }\quad x=y.\\ 1\ & \text{ if }\quad x\neq y. \end{cases}\quad\) Then, for \(x\in X\) and \(r>0\) we have \[B_{r}(x):=\begin{cases} \{x\}\ & \text{ if }\quad r\leq1.\\ X\ & \text{ if }\quad r>1. \end{cases}\] Therefore, (by Lemma [lem:open-balls-are-open]) every singleton \(\{x\}\) is an open set with respect to the discrete metric. Moreover, the complement \(X\setminus\{x\}\) is also open, since for any \(y\in X\setminus\{x\}\) (that is, any \(y\neq x\) in \(X\)) and any \(r<1\) the open ball \(B_{r}(y)=\{y\}\) is contained in \(X\setminus\{x\}\). Thus, all balls are clopen with respect to the discrete metric! In fact, any subset \(Y\subseteq X\) of a discrete metric space is clopen!

3. Sets don’t have to be either open or closed. For example, \([0,1)\) is neither open nor closed in \(\mathbb{R}\) (with respect to the standard metric \(|\,.\,|\)) - simply check the point \(x=0\) in \([0,1)\) and the point \(x=1\) in the complement \((-\infty,0)\cup[1,\infty)\). However, recall that any subset of a metric space is itself a metric space (w.r.t the same metric) - the subspace metric. Thus the pair \(([0,1),|\,.\,|)\) is a metric space - but then (by the remark after Definition [def:open-and-closed-sets]) the set \([0,1)\) is open!

   Key: Open and closed sets are really relative notions, depending on the ambient space (as well as the metric).

   Notation: When we say a subset of \(\mathbb{R}\) or \(\mathbb{R}^{n}\) or \(\mathbb{C}\) are open/closed, we will mean with respect to the standard norms \(|\,.\,|\) and \(\|\,.\,\|_{n}\) and \(|\,.\,|\) respectively. Most sets we encounter do not simply look like open/closed balls, so it will be useful to have rules for union and intersection:

We have the following remarks about open and closed sets.

• An infinite intersection of open sets is not necessarily open (see Sheet 2). Similarly, an infinite union of closed sets is not necessarily closed: e.g., the union of closed intervals in \(\mathbb{R}\), \[\bigcup_{i=1}^{\infty}\left[\frac{1}{i},\:1-\frac{1}{i}\right]\,=\,(0,1)\quad\text{ is open in }\mathbb{R}.\]

• The next generalisation of a metric space you will encounter (see Topology III) is called a Topological space \(\mathcal{T}\). There, the only stipulations are the existence of open sets such that \[(i)\:\emptyset\text{ and }\mathcal{T}\text{ are open };\quad\quad(ii)\:\text{Lemma 2.8 holds.} %\ref{lem:metrics-give-topology} holds.} %(open balls are open).}\] We have a hierarchy: \[\text{Inner-product space}\,\Longrightarrow\,\text{Normed space}\,\Longrightarrow\,\text{Metric space}\,\Longrightarrow\,\text{Topological space}.\]

• Why have we been looking at examples in \(\mathbb{R}^{2}\) rather than \(\mathbb{C}\)? It turns out that both these spaces are ‘topologically equivalent’, that is, they have the same open sets - this is obvious since the complex modulus is essentially just the Euclidean norm on \(\mathbb{R}^{2}\).

As we have seen, some sets are neither open nor closed. It will be useful to ask what the largest possible open set is inside a given set. Similarly, what is the smallest closed set containing a given set?

• Don’t confuse closure with conjugation! Closure concerns sets in any metric space, conjugation concerns points in \(\mathbb{C}\).

• Clearly the interior and exterior are open and clearly the boundary is closed. The closure is also closed (see sheet 2) - in fact, it is often defined more simply as \(\bar{A}:=X\setminus(X\setminus A)^{0}\), from which the closedness is obvious - it is the complement of an open set.

• The boundary matches our naive notion. Broadly speaking, the interior of a set consists of all the points that are not on its ‘edge’, and to form the closure of a set you simply add all the missing edge points. Indeed, we have the following additional properties of a subset \(A\subset X\) (see sheet 2): \[\begin{aligned} (a) & \:A\text{ is open } & \iff & \:\:\partial A\cap A=\emptyset & \iff & \:A=A^{0}; & \text{In fact } & \quad A^{0}={\displaystyle \bigcup_{\substack{U\subseteq A\\ U\text{ open} } }U;}\\ (b) & \:A\text{ is closed } & \iff & \quad\partial A\subseteq A & \iff & \:A=\bar{A}; & \text{In fact } & \quad\bar{A}={\displaystyle \bigcap_{\substack{A\subseteq F\\ F\text{ closed} } }F;\ }\end{aligned}\] That is, the interior \(A^{0}\) is the largest open set contained in \(A\) and the closure \(\bar{A}\) is the smallest closed set containing \(A\). Convince yourself that all the definitions reflect your intuitive notions, say, for the plane!

• In \(\mathbb{R}^{n}\) or \(\mathbb{C}^{n}\) for simple sets we only have to replace strict inequality with equality (or vice versa) to obtain the closure (or interior). For example, for \(A = \{ z \in \mathbb{C} : 1 < \vert z \vert \leq 3\}\), we have \(A^{0} = \{ z \in \mathbb{C} : 1 < \vert z \vert < 3\}\), \(\bar{A} = \{ z \in \mathbb{C} : 1 \leq \vert z \vert \leq 3\}\), and \(\partial A = \{ z \in \mathbb{C} : \vert z \vert =1 \} \cup \{ z \in \mathbb{C} : \vert z \vert =3\}\).

  Similarly \[\overline{\{z\in\mathbb{C}:\,1<\mathrm{Re}(z)\leq3,|\mathrm{Im}(z)|<1\}}=\{z\in\mathbb{C}:\,1\leq\mathrm{Re}(z)\leq3,|\mathrm{Im}(z)|\leq1\}\] and \[\{z\in\mathbb{C}:\,1<\mathrm{Re}(z)\leq3,|\mathrm{Im}(z)|<1\}^{0}=\{z\in\mathbb{C}:\,1<\mathrm{Re}(z)<3,|\mathrm{Im}(z)|<1\}.\] In fact, \(\overline{B_{r}(x)}=\bar{B}_{r}(x)\) for any ball in \(\mathbb{R}^{n}\) or \(\mathbb{C}^{n}\).

  However, this is not true in every metric space (see sheet 2) - there are metric spaces for which \(\overline{B_{r}(x)}\neq\bar{B}_{r}(x)\); that is, the smallest closed set containing the open ball \(B_{r}(x)\) is not the closed ball \(\bar{B}_{r}(x)\)! [Hint: what if the open ball is already closed!?]

Convergent sequences (example)

As mentioned, the chordal metric on \(\hat{\mathbb{C}}\) is \(d(z,w)=\|f(z)-f(w)\|_{2}\), where \(\|\,.\,\|_{2}\) is the Euclidean norm on \(\mathbb{R}^{3}\) and \(f\) is the inverse Stereographic projection given by \[f(z)=\left(\frac{2\mathrm{Re}(z)}{|z|^{2}+1},\:\frac{2\mathrm{Im}(z)}{|z|^{2}+1},\:\frac{|z|^{2}-1}{|z|^{2}+1}\right).\] Show that with respect to this metric the sequence \(\{ki\}_{k\in\mathbb{N}}\) in \(\hat{\mathbb{C}}\) converges to \(\infty\in\hat{\mathbb{C}}\).

Since \(|ki|=k\) and \(f(\infty)=(0,0,1)\) we have \[\begin{aligned} d(ki,\infty)&=\|f(ki)-f(\infty)\|_{2} = \left\Vert \left(0,\frac{2k}{k^{2}+1},\frac{k^{2}-1}{k^{2}+1}\right)-(0,0,1)\right\Vert _{2}\\ & = \left\Vert \left(0,\frac{2k}{k^{2}+1},-\frac{2}{k^{2}+1}\right)\right\Vert _{2}\\ & = \sqrt{\left(\frac{2k}{k^{2}+1}\right)^{2}+\left(\frac{-2}{k^{2}+1}\right)^{2}}\quad\longrightarrow\quad0\quad\text{ as }k\rightarrow\infty.\end{aligned}\] Thus, the sequence indeed converges to \(\infty\). This is quite an odd notion as we are used to saying sequences ‘diverge’ if they tend to infinity. The key is that convergence depends on the metric being used.

Limits in \(\mathbb{C}\) with the standard metric (Very important!).

The above definition of limit in a metric space also gives us an example of limits in \(\mathbb{C}\) with the standard metric. This says that if \(z_{n}\) is a sequence of complex numbers, then \(\lim_{n\to\infty}z_{n}=z\) if and only if

‘for all \(\epsilon>0\), there exists \(N>0\) such that for all \(n\geq N\), \(|z_{n}-z|<\epsilon\)’.

Note this is the same definition as in Analysis I, but replacing the absolute value on the real line by the modulus on \(\mathbb{C}\). Importantly, by the same proofs as in Analysis I, limits in the complex plane follow the COLT rules.

Furthermore there is a very important link between convergence in the complex plane and real convergence (see Analysis I for the proof). Let \(\{z_{n}\}_{n\in\mathbb{N}}\) be a sequence of complex numbers \(z_{n}=x_{n}+iy_{n}\). Then, for any fixed \(z_{0}=x_{0}+iy_{0}\in\mathbb{C}\) we have \[\lim_{n\rightarrow\infty}z_{n}=z_{0}\quad\iff\quad\lim_{n\rightarrow\infty}x_{n}=x_{0}\:\:\text{ and }\:\:\lim_{n\rightarrow\infty}y_{n}=y_{0}.\] In other words, the sequence \(\{z_{n}\}\) converges iff the real sequences \(\{\mathrm{Re}(z_{n})\}\) and \(\{\mathrm{Im}(z_{n})\}\) converge (see Sheet 3).

Let’s return to our general setting of metric spaces and prove some properties of limits.

1. A sequence can have at most one limit.

2. We have \[\lim_{n\to\infty}x_{n}=x\quad\Longleftrightarrow\quad\forall\text{ open }U\text{ with }x\in U,\:\exists\,N\in\mathbb{N}\text{ such that }\:\forall n>N\quad x_{n}\in U.\] Hence the notion of a limit in a metric space can be stated in terms only of its open sets.

We can now define what it means for a function between two metric spaces to be continuous (this will incorporate many of the functions we have already encountered; e.g., \(f:\mathbb{C}\to\mathbb{C};\:\mathbb{R}\to\mathbb{C};\) or \(\mathbb{C}\to\mathbb{R}\)).

Examples of continuous functions on the complex plane (with the standard metric)

• The identity function is continuous.

• Constant functions are continuous.

• The functions \(\mathrm{Re},\mathrm{Im}:\mathbb{C}\to\mathbb{R}\) are continuous.

• The complex conjugation \(z\mapsto\bar{z}\) is continuous as a map from \(\mathbb{C}\to\mathbb{C}\).

• The modulus function \(z\mapsto|z|\) is continuous as a map from \(\mathbb{C}\to\mathbb{R}\).

• All of \(\exp,\sin,\cos,\sinh\) and \(\cosh\) are continuous on \(\mathbb{C}\), as are all polynomials.

• If \(\arg\) is the choice of argument function with values in \((\theta_{1},\theta_{2}]\) then \(\arg\) is continuous on \(\mathbb{C}\backslash R_{_{\theta_{1}}}\) (recall \(R_{\theta_{1}}\) is the ray with angle \(\theta_{1}\))

• If \(\log\) is a branch of \(\log\) corresponding to an argument function as above, then \(\log\) is continuous on \(\mathbb{C}\backslash R_{\theta_{1}}\).

As with limits, it will be useful to restate continuity in terms of open sets. First, recall that for any function \(f:X_{1}\to X_{2}\) and any set \(U\subseteq X_{2}\) we define the preimage \(f^{-1}(U)\) of \(U\) under \(f\) by \(f^{-1}(U):=\{x\in X_{1}:f(x)\in U\}\).

As with limits, it turned out that continuity depends only upon the open sets in the respective metric spaces. This means we can use the continuity of known functions to prove the openness of very complicated sets.

Showing sets are open using continuity (examples)

• Show the following set is open: \[U=\{(x,y)\in\mathbb{R}^{2}:(x^{2}+y^{2})\sin^{3}(\sqrt{x^{2}+7})>2\}.\] Well, the function \[f:\mathbb{R}^{2}\to\mathbb{R}:\:(x,y)\mapsto(x^{2}+y^{2})\sin^{3}(\sqrt{x^{2}+7})\] is continuous by Lemma [lem:combining-continuous-function], because it is the product/composition of real valued continuous functions. Moreover, \[U=\{(x,y)\in\mathbb{R}^{2}:f((x,y))>2\}\:=\:f^{-1}((2,\infty)).\] Since \((2,\infty)\) is open in \(\mathbb{R}\) (see sheet 2), the set \(U\) is the preimage of an open set under a continuous map and by Theorem [thm:continuity-open-sets] it is therefore open.

• We can actually do a little more using the following useful properties of the preimage (from Analysis I):

  Useful properties of preimage

  • \(f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)\).

  • \(f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)\).

  • \(f^{-1}(A\setminus B)=f^{-1}(A)\setminus f^{-1}(B)\).

  Show the following set is open: \[U=\{(x,y)\in\mathbb{R}^{2}:xy>1,\:x^{2}+y^{2}>3\}.\] Both \[f(x,y)=xy\quad\text{and}\quad g(x,y)=x^{2}+y^{2}\] are continuous as functions \(\mathbb{R}^{2}\to\mathbb{R}\), and \(U=f^{-1}((1,\infty))\:\cap\:g^{-1}((3,\infty))\).

  Since \(f\) and \(g\) are continuous and both \((1,\infty)\) and \((3,\infty)\) are open in \(\mathbb{R}\), the preimages \(f^{-1}((1,\infty))\) and \(g^{-1}((3,\infty))\) are open (by Theorem [thm:continuity-open-sets]). By Lemma [lem:metrics-give-topology] we have that \(U\) is open, since it is the intersection of two open sets.

Showing functions are not continuous using open sets (example)

• We can also use openness to prove a function is not continuous. Indeed, for \(f :X_1 \to X_2\), if there exists an open set \(U\) in \(X_2\) such that \(f^{-1}(U)\) is not open in \(X_1\), then \(f\) is not continuous. For example, the function \(f:\mathbb{R}^{2}\to\mathbb{R}\) defined by \[f(x,y)=\begin{cases} \frac{xy}{x^{2}+y^{2}}, & \text{ if }(x,y)\neq0,\\ 0 & \text{ otherwise},\ \end{cases}\quad\] isn’t continuous at \((0,0)\). Why? Consider the preimage \(f^{-1}((-\epsilon,\epsilon))\). Claim: This preimage is not open for \(\epsilon\) sufficiently small.

  First note that the preimage in question contains \((0,0)\) since \(f((0,0))=0\). To show the preimage is not open it is enough to show that any open ball in \(\mathbb{R}^{2}\) centred at \((0,0)\) is not contained in \(f^{-1}((-\epsilon,\epsilon))\): Let \(\epsilon<1/4\), say, and for any \(\delta>0\) consider the ball \(B_{\delta}((0,0))\) centred at \((0,0)\). The point \((\delta/2,\delta/2)\) is in \(B_{\delta}((0,0))\) since \(\|(\frac{\delta}{2},\frac{\delta}{2})-(0,0)\|_{2}=\|(\frac{\delta}{2},\frac{\delta}{2})\|_{2}=\sqrt{\frac{\delta^{2}}{4}+\frac{\delta^{2}}{4}}=\frac{\delta}{\sqrt{2}}<\delta\). But \[f\left(\left(\frac{\delta}{2},\frac{\delta}{2}\right)\right)\quad=\quad\frac{\frac{\delta}{2}\cdot\frac{\delta}{2}}{\frac{\delta^{2}}{4}+\frac{\delta^{2}}{4}}\quad=\quad\frac{1}{2}\quad>\quad\epsilon,\] so \((\delta/2,\delta/2)\) is not in \(f^{-1}((-\epsilon,\epsilon))\). Thus, for any \(\delta>0\) the ball \(B_{\delta}((0,0))\) is not contained in \(f^{-1}((-\epsilon,\epsilon))\) and so this preimage is not open.

  Since \((-\epsilon,\epsilon)\) is open in \(\mathbb{R}\) it follows from Theorem [thm:continuity-open-sets] that \(f\) is not continuous.

Why preimages?

Note that the use of preimages in Theorem [thm:continuity-open-sets], rather than images, is important. The same result is not true of images. E.g., the function \(f(z)=|z|\) is continuous as a function \(\mathbb{C}\to\mathbb{R}\), but it maps an open set in the complex plane \(f:\mathbb{D}\to[0,1)\) to an interval that is neither open nor closed in \(\mathbb{R}\).

Note that this \(f\) is actually a bijection from \(\mathbb{R}_{\geq 0}\) to \(\mathbb{R}_{\geq0}\), and in \(\mathbb{R}_{\geq0}\) the interval \([0,1)\) is open! So, is the problem that we need the function to be bijective? No. For example, consider the metric spaces \(X_{1}=[0,1)\cup[2,3]\) and \(X_{2}=[0,2]\) with the usual (subspace) metric coming from the absolute value on \(\mathbb{R}\). Define \[f:X_{1}\to X_{2}:\:x\mapsto\begin{cases} x, & \text{ if }x\in[0,1).\\ x-1, & \text{ if }x\in[2,3].\ \end{cases}\] It is easy to check that \(f\) is a bijection and is continuous on its domain: (Continuity is trivial for \(x\neq2\). For \(x=2\), pick \(\epsilon>0\), then for any \(0<\delta<1\) we have \(B_{\delta}(2)=[2,2+\delta)\). Note that \(f(2)=1\) and so \(B_{\epsilon}(f(2))=(1-\epsilon,1+\epsilon)\). To show \(f\) is continuous we must therefore find a \(\delta\) so that \(f(x)\in(1-\epsilon,1+\epsilon)\) if \(x\in[2,2+\delta)\). Simply pick any \(\delta<\epsilon\), for then: \[x\in[2,2+\delta)\quad\Rightarrow\quad f(x)=x-1\in[1,1+\delta)\subset(1-\delta,1+\delta)\subset(1-\epsilon,1+\epsilon),\] as required.) But, the set \([2,3]\) is open in \(X_{1}\) (see sheet 3) and its image \(f([2,3])=[1,2]\) is not open in \(X_{2}\).

Thus, we genuinely do need to use preimages. When can we use images of continuous functions to preserve properties of the sets in question? When can we find the maximum/minimum value taken by a function on a set? It turns out a key concept is that of compactness.

Differentiating complex functions from first principles (examples)

1. Rules for differentiating polynomials are the same. For example, consider \(f(z)=z^{2}\) on \(\mathbb{C}\). For any \(z\in\mathbb{C}\) we have \[\lim_{h\to0}\frac{(z+h)^{2}-z^{2}}{h}=\lim_{h\to0}\frac{z^{2}+2hz+h^{2}-z^{2}}{h}=\lim_{h\to0}(2z+h)=2z.\] Thus, \(f\) is differentiable on \(\mathbb{C}\) and \(f'(z)=2z\) as expected.

2. Consider \(f(z)=\bar{z}\). For it to be differentiable we must obtain the same limit from every direction. But, considering limits from the purely real and purely imaginary directions, for every \(z\in\mathbb{C}\) we have \[\lim_{\substack{h\rightarrow0\\ h\in\mathbb{R} } }\frac{\overline{z+h}-\bar{z}}{h}=\lim_{\substack{h\rightarrow0\\ h\in\mathbb{R} } }\frac{h}{h}=1,\] yet \[\lim_{\substack{h\rightarrow0\\ h\in\mathbb{R} } }\frac{\overline{z+ih}-\bar{z}}{ih}=\lim_{\substack{h\rightarrow0\\ h\in\mathbb{R} } }\frac{-ih}{ih}=-1.\] Since \(z\) was arbitrary this shows \(f\) is not differentiable anywhere.

3. As in the real case, sums/products/quotients of complex differentiable functions are complex differentiable where defined (e.g., all polynomials/rational functions). In particular, the product and quotient rules hold for complex derivatives.

4. Composition of differentiable functions are complex differentiable where defined. In particular, the chain rule holds for complex derivatives. The proofs of \(3.\) and \(4.\) here are almost identical to those from Analysis I, so are excluded.

5. Generally, non-constant purely real/imaginary functions are not complex differentiable ; e.g., \(\mathrm{Re}(z),\mathrm{Im}(z),|z|\) are nowhere differentiable as functions from \(\mathbb{C}\to\mathbb{C}\).

Determining complex differentiability via Cauchy-Riemann (examples)

1. Let \(f(z)=\exp(z)=e^{x}\cos y+ie^{x}\sin y\). Then \[u_{x}=e^{x}\cos y,\quad v_{y}=e^{x}\cos y,\quad u_{y}=-e^{x}\sin y,\quad v_{x}=e^{x}\sin y.\] All these functions are continuous as real functions (see Calculus I/AMV II) and the C-R equations hold. Thus, by Theorem [thm:Cauchy-Riemann-implies-complex-differentiability], \(\exp\) is differentiable everywhere in \(\mathbb{C}\) and by Proposition [prop:complex-diff-implies-CR] \[\exp'(z)=u_{x}+iv_{x}=e^{x}\cos y+ie^{x}\sin y=\exp(z).\]

2. By the chain rule, \(f(z)=e^{iz}\) is differentiable and \(f'(z)=ie^{iz}\). Since they are just sums of \(\exp\), all of the functions \(\sin,\cos,\sinh,\cosh\) are differentiable everywhere in \(\mathbb{C}\) and you can verify \[\begin{aligned} \sin'(z)=\cos z,\quad\cos'(z)=-\sin z,\quad\sinh'(z)=\cosh z,\quad\cosh'(z)=\sinh z.\end{aligned}\] Similarly, all polynomials/rational functions are differentiable with the usual formulae: e.g., for \(a_{0}\ldots a_{n}\) complex; \[(a_{n}z^{n}+\cdots+a_{2}z^{2}+a_{1}z+a_{0})'=na_{n}z^{n-1}+\cdots+2a_{2}z+a_{1}.\] For the branch of log corresponding to arguments in \((\theta_{1},\theta_{2}]\), the function \(\log(z)\) is differentiable in \(\mathbb{C}\setminus R_{\theta_{1}}\), i.e. at all points outside the branch cut. At these points, the derivative is given by \(\log'(z)=1/z\) (see Sheet 4 Q7 where this is proved for the principal branch).

3. \(f(z)=\sin(z)/z^{2}\) is differentiable everywhere except \(z=0\), since it is the quotient of two complex differentiable functions (and is not defined when \(z^{2}=0\)). Furthermore, \[\left[\sin(z)/z^{2}\right]'=\frac{(\cos z)(z^{2})-(\sin z)(2z)}{(z^{2})^{2}}=\frac{z\cos z-2\sin z}{z^{3}}.\]

4. Let \(f(z)=f(x+iy)=(x^{3}+3x^{2}y-y^{3}-x^{2}-2y^{2})+i(-x^{3}+3xy^{2}-y^{3}+4xy+3y)\). Then, \[\begin{aligned} u_{x}=3x^{2}+6xy-2x,\quad\quad & v_{y}=6xy-3y^{2}+4x+3,\\ u_{y}=3x^{2}-3y^{2}-4y,\quad\quad & v_{x}=-3x^{2}+3y^{2}+4y.\end{aligned}\] Thus, the partial derivatives exist and are continuous everywhere. We have \(u_{y}=-v_{x}\) everywhere, but \(u_{x}=v_{y}\) if and only if \(3x^{2}-2x=-3y^{2}+4x+3\); that is, when \((x-1)^{2}+y^{2}=2\). So, \(f\) is differentiable only on the circle of radius \(\sqrt{2}\) centred at \(1\) in the complex plane!

Holomorphicity

In example \(4.\) above, the function is only differentiable on a 1-dimensional subset of \(\mathbb{C}\). This is a similar situation to a real function on \(\mathbb{R}\) being differentiable only at a single point - not a very interesting function to work with from an analytical perspective. It will be useful for us to consider functions that are differentiable on genuine \(2\)-dimensional sets in \(\mathbb{C}\), for this will allow us to (later) express the functions using Taylor series. Such functions will turn out to have some quite remarkable properties.

We say \(f\) is holomorphic at \(\boldsymbol{z_{0}\in U}\) if it is holomorphic on some open ball \(B_{\epsilon}(z_{0})\); (in other words, if there exists \(\epsilon>0\) such that \(f\) is complex differentiable at every point in \(B_{\epsilon}(z_{0})\)).

Showing sets are domains (examples)

1. We know both \(\mathbb{C}\) and \(B_{r}(z)\) (for any \(z\in\mathbb{C}\) and \(r>0\)) are open. They are also path-connected: Simply choose the line segment between \(a\) and \(b\) given by \(\gamma(t)=a+(b-a)t\). This is clearly a smooth path, so \(\mathbb{C}\) and \(B_{r}(z)\) are domains.

2. The set \(\mathbb{C}\setminus\mathbb{R}_{\leq0}\), on which the principal branches of \(\log\) and complex powers are defined, is path-connected: Choose arbitrary \(a,b\in\mathbb{C}\setminus\mathbb{R}_{\leq0}\). If \(a\) lies on the positive real axis then the line segment again works (since it avoids the negative real axis). Otherwise, we could try letting \[\gamma_{1}(t)=|a|e^{i\mathrm{Arg}(a)(1-t)},\quad\quad\gamma_{2}(t)=|a|+(b-|a|)t.\] The first path takes \(a\) to \(|a|\), avoiding \(\mathbb{R}_{\leq0}\) by tracing around the circle of radius \(|a|\) centred at the origin. The second joins \(|a|\) to \(b\) via a straight line. (You can reparametrize to make \(t\) run from \(0\) to \(1\) on the combined path if you like.) But the combined path running through \(\gamma_1\) then \(\gamma_2\) is not differentiable at the point where the paths \(\gamma_1\) and \(\gamma_2\) meet! One possibility for a differentiable path joining the points \(a\) and \(b\) would be the arc of the circle through \(a\) and \(b\) that does not cross \(\mathbb{R}_{\leq0}\).

   We know \(\mathbb{C}\setminus\mathbb{R}_{\leq0}\) is also open, so it is a domain.

3. The set \(\{z\in\mathbb{C}:|z|\neq1\}\) is not a domain. It is open, but it is not path-connected. Pick a point \(a\) with \(|a|<1\) and \(b\) with \(|b|>1\). We cannot draw a continuous path from \(a\) to \(b\) without crossing the circle \(|z|=1\); i.e., you cannot draw a continuous path between the points without leaving the set.

Previously we have seen that there is a chain rule for the composition of two complex differentiable functions. There is also the chain rule for the composition of a complex differentiable function and a smooth path:

Determining where maps are conformal (examples)

• Let \(f(z)=z^{2}\). Here, \(f'(z)=2z\) so \(f'(z)=0\iff z=0\) and so \(f\) is conformal on \(\mathbb{C}^{\ast}=\mathbb{C}\setminus\{0\}\) (since it is holomorphic everywhere). We don’t yet know whether \(f\) is conformal at the origin, but we can check: we know that \(f'(z)=2z\) so \(f'(0)=0\) and hence the action of \(f\) on tangent vectors at the origin is to multiply by \(0\), hence sending any tangent vector to zero.

• Let \(f(z)=\frac{2}{3}z^{3}+\frac{1}{2}(1-2i)z^{2}-iz+2-1\). This function is clearly holomorphic as it is just a polynomial. We have \[f'(z)=2z^{2}+(1-2i)z-i=(2z+1)(z-i),\] which has zeros at \(-\frac{1}{2}\) and \(i\), so \(f\) is conformal on \(\mathbb{C}\setminus\{-\frac{1}{2},i\}\). [Note, we can explicitly check what \(f\) does to tangent vectors at each point in this set: For example, at \(z=i/2\) we have \(f'(i/2)=(i+1)(-i/2)=(1-i)/2=\frac{1}{\sqrt{2}}e^{-i\pi/4}.\) Thus, at \(z=i/2\) the function \(f\) rotates tangent vectors by \(\mathrm{Arg}(f'(i/2))=-\pi/4\) degrees, then dilates by a factor of \(|f'(z)|=1/\sqrt{2}\).]

  Rather than having to then check two paths through \(z=1/2\) and \(z=i\) respectively, to determine whether \(f\) is conformal there, it will be convenient to have a complete description in terms of holomorphicity. The converse to Lemma [lem:A-holomorphic-map-is-conformal] holds:

Determining where maps are conformal (examples continued)

• Let \(f(z)=xy+iy^{2}\). Then \(u_{x}=y,v_{y}=2y,u_{y}=x,v_{x}=0\). So, the C-R equations only hold at \(y=x=0\). Hence \(f\) is not conformal at any point \(z_{0}\neq0\), since it cannot be complex differentiable at \(z_{0}\) (as C-R. equations fail to hold), and hence it is not conformal at \(z_{0}\) by Proposition [prop:conf-to-holo]. What happens for \(z_{0}=0\)? There, \(f'(0)=u_{x}(0,0)+iv_{x}(0,0)=0+i0=0\). So \(f\) cannot be conformal at \(0\) either by Proposition [prop:conf-to-holo]. Thus, \(f\) is nowhere conformal.

Visualising conformal maps

Here is a useful result, helpful in visualising the action of conformal maps:

How to visualise conformal maps (examples)

• Let \(f(z)=z^{2}=x^{2}-y^{2}+i2xy\) and consider the simple grid in the \((x,y)\)-plane made up of lines parallel to the real/imaginary axes, separated by distance \(1\).

  Given a line (not passing through the origin) in such a grid, say \(x=a\,(\neq0)\), the function \(f\) takes this line to the points \((a^{2}-y^{2},2ay)\) in the \((u,v)\)-plane. If \(u=a^{2}-y^{2}\) and \(v=2ay\), then \(v^{2}=4a^{2}y^{2}\) and so \(u=a^{2}-(\frac{v}{2a})^{2}\) and the image defines a parabola.

  Similarly, the line \(y=b\,(\neq0)\) is taken to the parabola \(u=(\frac{v}{2b})^{2}-b^{2}\) in the \((u,v)\)-plane. By sketching these on the same axes we see that the parabolas indeed cross at right angles.

• For the same example, consider the level curves \(u(x,y)=a\) and \(v(x,y)=b\) with \(a,b\neq0\). These trace out the curves \(y^{2}=x^{2}-a\) and \(y=\frac{b}{2x}\) respectively. By sketching these curves on a graph (for, say, \(u=\pm1,v=\pm1\)) we see that they are perpendicular. (They have to be, since \(f(z)\) is conformal on \(\mathbb{C}-\{0\}\) and they map to perpendicular straight lines.)

Finding biholomorphisms (examples)

1. The function \(\exp:\mathbb{C}\to\mathbb{C}^{\ast}\) is not biholomorphic since it is not injective (e.g., \(e^{z}=e^{z+2n\pi i}\)). How can we make it biholomorphic? We simply need to restrict it to a smaller domain where it will be injective. It is not too hard to check that \[\exp(z_{1})=\exp(z_{2})\] if and only if \(\mathrm{Re}(z_{1})=\mathrm{Re(}z_{2})\) and \(\mathrm{Im}(z_{1})-\mathrm{Im}(z_{2})\in2\pi\mathbb{Z}\). So if we rule out different \(z_{1}\) and \(z_{2}\) whose imaginary parts differ by an non-zero integer multiple of \(2\pi\) then we might be ok. We can do this by setting \[D:=\{z\in\mathbb{C}:\mathrm{Im}(z)\in(-\pi,\pi)\:\}.\] If \(z=x+iy\in D\) then \(\exp(z)=e^{x}e^{iy}\) so the principal argument of \(\exp(z)\) is never equal to \(\pi\). On the other hand, it is not hard to check that \(\exp\) maps \(D\) onto \(\mathbb{C}-\mathbb{R}_{\leq0}\). Moreover, by our choice of \(D\), \(\exp\) is injective on \(D\). So \(\exp\) is a bijection between \(D\) and \(\mathbb{C}-\mathbb{R}_{\leq0}\). Therefore it has an inverse function, and we already know what it is: the principal branch of \(\mathrm{Log}\). The function \(\mathrm{Log}\) maps \(\mathbb{C}-\mathbb{R}_{\leq0}\) to \(D\) and is differentiable at every point of \(\mathbb{C}-\mathbb{R}_{\leq0}\), so it is holomorphic on \(\mathbb{C}-\mathbb{R}_{\leq0}\). In summary, when restricted to \(D\), \(\exp\) is biholomorphic and hence \(D\) and \(\mathbb{C}-\mathbb{R}_{\leq0}\) are biholomorphic.

2. Let \(f(z)=z^{2}\). We have seen that this function is holomorphic on \(\mathbb{C}\). However, as in the previous example, it is not injective since e.g. \(f(1)=f(-1)=1\). Note that \(f(z_{1})=f(z_{2})\) with \(z_{1}\neq z_{2}\) if and only if \(z_{1}=\pm z_{2}\), so we may be able to create a biholomorphic function by restricting \(f\) to a domain that never simultaneously contains \(z\) and \(-z\). Such a domain is the right half plane \(\mathbb{H}_{R}:=\{z\in\mathbb{C}:\mathrm{Re}(z)>0\}\). We saw in a previous example that \(f\) maps \(\mathbb{H}_{R}\) to \(\mathbb{C}-\mathbb{R}_{\leq0}\). In fact this is a bijection, with inverse given by \[f^{-1}:\mathbb{C}-\mathbb{R}_{\leq0}\to\mathbb{H}_{R},\quad f^{-1}(z):=\exp(\frac{1}{2}\mathrm{Log}(z)).\] Thus \(f^{-1}\) is holomorphic, since it is a composition of holomorphic functions, and hence \(f\) gives a biholomorphism \(f:\mathbb{H}_{R}\xrightarrow{\sim}\mathbb{C}-\mathbb{R}_{\leq0}.\)

3. The ‘affine’ linear maps \(z\mapsto az+b\) (for \(a\in\mathbb{C}^{\ast},b\in\mathbb{C}\)) are biholomorphic \(\mathbb{C}\xrightarrow{\sim}\mathbb{C}\).

As expected we can compose biholomorphic maps to construct new ones.

Finding a Möbius transformation from three points (example)

Find the unique Möbius transformation \(f:\hat{\mathbb{C}}\to\hat{\mathbb{C}}\) mapping the points \(\{1,-1,i\}\) to the points \(\{0,\infty,1\}\). Our method is simply to rearrange \((\ast)\): Notice that \[\lim_{|w|\to\infty}\frac{f(z)-w}{w_{1}-w}\:=\:\lim_{|w|\to\infty}\frac{\frac{f(z)}{w}-1}{\frac{w_{1}}{w}-1}\:=\:\frac{-1}{-1}\:=\:1,\] so \((\ast)\) reduces to \[\begin{aligned} 1\cdot\frac{w_{1}-w_{3}}{f(z)-w_{3}}\quad &=\quad\left(\frac{z-z_{2}}{z-z_{3}}\right)\left(\frac{z_{1}-z_{3}}{z_{1}-z_{2}}\right)\\ \iff \frac{0-1}{f(z)-1}\quad & =\quad\left(\frac{z-(-1)}{z-i}\right)\left(\frac{1-i}{1-(-1)}\right)\\ \iff f(z)-1\quad & =\quad\frac{-2(z-i)}{(z+1)(1-i)}\\ \iff f(z)\quad & =\quad\frac{(-1-i)z+(1+i)}{(1-i)z+(1-i)}\\ \iff f(z)\quad & =\quad\left(\frac{-1-i}{1-i}\right)\left(\frac{z-1}{z+1}\right)\\ \iff f(z)\quad & =\quad\frac{-iz+i}{z+1}.\end{aligned}\]

Möbius transformations methods

This leads us to a general strategy to find a Möbius transformation from how it acts on three points: Simply notice that since the cross ratio is preserved, \[\left(\frac{f(z)-w_{2}}{f(z)-w_{3}}\right)\left(\frac{w_{1}-w_{3}}{w_{1}-w_{2}}\right)\quad=\quad\left(\frac{z-z_{2}}{z-z_{3}}\right)\left(\frac{z_{1}-z_{3}}{z_{1}-z_{2}}\right),\] and to find \(f\) we need only rearrange this equation.

There is also a general strategy to find the image of a region \(D\) under a Möbius transformation \(M_T\):

1. Find the image \(M_{T}(\partial D)\) of the boundary \(\partial D\).

2. Find the image \(M_{T}(z_{0})\) of a point \(z_{0}\in D\) in the interior.

3. The region \(D'\) bounded by \(M_{T}(\partial D)\) and containing \(M_{T}(z_{0})\) is precisely the image of \(D\) under \(M_{T}\), and \[M_{T}:D\:\xrightarrow{\sim}\:D'=M_{T}(D).\]

Finding the image of regions under Möbius transformations (example)

Find the image of the unit disc \(\mathbb{D}:=B_{1}(0)=\{w\in\mathbb{C}:|w|<1\}\) under the Möbius transformation corresponding to the matrix \(T=\left(\begin{array}{cc} 2+2i & -2-6i\\ 1 & -1-2i \end{array}\right)\). To determine the image we look at what happens to the boundary: we first find the image of the unit circle centred at the origin. Consider what happens to the four points on the unit circle, \(1,i,-1,-i\) under the map \(M_{T}(z)=\frac{(2+2i)z-(2+6i)}{z-(1+2i)}\). Under \(M_{T}\): \[\begin{aligned} & 1 & \mapsto\frac{(2+2i)-(2+6i)}{1-(1+2i)}=\frac{-4i}{-2i}=2;\\ & i & \mapsto\frac{(2+2i)i-(2+6i)}{i-(1+2i)}=\frac{-4-4i}{-1-i}=4;\:\\ & -1 & \mapsto\frac{-(2+2i)-(2+6i)}{-1-(1+2i)}=\frac{-4-8i}{-2-2i}\end{aligned}\] By Proposition [prop:circlines] the unit circle must be mapped to a circle, and the three calculations above tell us that the image of the unit circle must be the circle centred at \(z=3\) of radius \(1\). Note, the action of \(M_{T}\) is not something so simple as a rotation; for example we have \[-i\mapsto\frac{-(2+2i)i-(2+6i)}{-i-(1+2i)}=\frac{-8i}{-1-3i}=\frac{12+4i}{5},\quad\text{which is near the image of -1 !!}\]

What about the interior of this circle? Let’s pick a point in the unit disc and see where it is taken. The point \(z=0\) is an obvious choice. We have \[M_{T}(0)=\frac{(2+2i)0-(2+6i)}{0-(1+2i)}=\frac{2+6i}{1+2i}=\frac{(2+6i)(1-2i)}{5}=\frac{14+2i}{5},\] which lies inside this circle in question. Thus, by continuity \(M_{T}\) maps \(\mathbb{D}\) to the interior of the circle centred at \(z=3\) of radius \(1\) (it cannot map another point \(z_{0}\in\mathbb{D}\) to somewhere outside this circle as the image of the path from \(0\) to \(z_{0}\) would have to cross the boundary, but Möbius transformations are conformal). Continuity (and the existence of an inverse) tell us the map must be onto and so \[M_{T}:\mathbb{D}\:\xrightarrow{\sim}\:\{z\in\mathbb{C}:|z-3|<1\}.\]

Finding automorphisms (example)

Find a Möbius transformation \(f\) from the closed unit disc onto the closed unit disc taking \(\frac{i}{2}\) to \(0\) and \(-i\) to \(1\). Since \(\frac{i}{2}\mapsto0\), by Corollary D2D* with \(z_{0}=\frac{i}{2}\) we have \[f(z)=e^{i\theta}\left(\frac{z-\frac{i}{2}}{-\frac{i}{2}z-1}\right)=e^{i\theta}\left(\frac{2z-i}{-iz-2}\right)\] for some \(\theta\). Since \(-i\mapsto1\), we have \[1=e^{i\theta}\left(\frac{-2i-i}{-1-2}\right)\quad\iff\quad1=e^{i\theta}i\quad\iff\quad e^{i\theta}=-i.\] Thus \[f(z)=-i\left(\frac{2z-i}{-iz-2}\right)=i\left(\frac{2z-i}{iz+2}\right)=\frac{2z-i}{z-2i}\quad(\text{multiplying top and bottom by }-i).\]

Finding the image of geometric shapes under Möbius transformation (example)

Let \(F\) be the geometric figure made up of a line segment from \(0\) to \(-1\), a clockwise circular arc (tracing out the unit circle) from \(-1\) to \(i\), then a line segment from \(i\) to \(0\). Find the image of \(F\) under the Cayley Map \(M_{C}(z)=\frac{z-i}{z+i}\).

Each section is a segment of a circle or a line, so must be taken to a segment of a circle or a line by Proposition [prop:circlines]. First, let us check where the three ‘vertices’ go. We have \[0\mapsto\frac{-i}{i}=-1;\quad-1\mapsto\frac{-1-i}{-1+i}=\frac{(1+i)^{2}}{2}=i;\quad i\mapsto\frac{i-i}{i+i}=0.\] Let’s look at where the line segment from \(0\) to \(i\) goes: \[\frac{i}{2}\mapsto\frac{\frac{i}{2}-i}{\frac{i}{2}+i}\,=\,\frac{-3i}{i}\,=\,-\frac{1}{3},\] so the line segment from \(0\) to \(i\) must map to the circle/line from \(-1\) to \(0\), passing through \(-\frac{1}{3}\); that is, it is the line segment from \(-1\) to \(0\) on the real axis.

Now, if we wanted we could do the same for the other two sections; pick a point on each remaining line/circular arc and see where it maps to, thus determining whether each image is a line segment or a circular arc. However, there is a much quicker method: We know Mob trans are conformal and so preserve angles/orientation - so tracing round the shape we must have the same angles in the image of the figure. So, travelling from \(M_{C}(0)=-1\), when we reach \(M_{C}(i)=0\) we must turn anticlockwise \(\pi/2\) degrees and head towards \(M_{C}(-1)=i\). Thus, the next section must simply be the straight line from \(0\) to \(i\). Similarly, the final section must be a circular arc from \(i\) back to \(-1\). Thus, the image is just \(F\). Actually, letting \(D=\{z\in\mathbb{D}:\pi/2<\mathrm{Arg}(z)<\pi\}\) be the interior of \(F\), one can check that \(D\) maps to \(D\) and and so \(M_{C^{-1}}\) (and therefore \(M_{C}\)) is actually in \(\mathrm{Aut}(D)\).

The problem with pointwise convergence (examples)

1. Consider the sequence of functions \(f_{n}=x^{n}\) on \([0,1]\). It is easy to see that sequence is pointwise convergent on \([0,1]\) with limit \[f(x)=\begin{cases} 0, & \quad\text{if }x<1.\\ 1, & \quad\text{if }x=1.\ \end{cases}\] This is a sequence of continuous, differentiable functions on a compact set, but the limit function is not continuous!

2. We encounter this issue again in the complex plane. Let \(f_{n}:\mathbb{C}\to\mathbb{C}:z\mapsto z^{n}\). We split into three cases.

   • Pick \(z\in\mathbb{D}\) and \(\epsilon>0\). Then, we can certainly find \(N\in\mathbb{N}\) such that \(|z|^{N}<\epsilon\) (for example, take any \(N>\log\epsilon/\log|z|\)). Thus, for every \(n>N\) we have \[|f_{n}(z)-0|=|z|^{n}<|z|^{N}<\epsilon,\] and so \(\lim_{n\to\infty}f_{n}(x)=0\) in \(\mathbb{D}\).

   • When \(|z|=1\) the point \(z\) rotates around the unit circle \(\partial\mathbb{D}\) by \(\mathrm{Arg}(z)\) every iteration. For any \(z\neq1\) this sequence clearly doesn’t converge, but for \(z=1\) we have \(\lim_{n\to\infty}f_{n}(z)=\lim_{n\to\infty}1=1\).

   • Here, the value \(|z|^{n}\) is unbounded, so the limit does not exist.

   To conclude, the sequence \(f_{n}(z)\) is not pointwise convergent on \(\mathbb{C}\), however, notice that it is pointwise convergent on \(\mathbb{D}\cup\{1\}\) with limit function \[f(z)=\begin{cases} 0, & \quad\text{if }z\in\mathbb{D}.\\ 1, & \quad\text{if }z=1.\ \end{cases}\] Again, this is not continuous.

3. Was the problem that the region where the function converges wasn’t open? From the previous examples it seems like this could be the case (it was fine on the interior of balls). No! For example, the sequence of continuous functions \(f_{n}(x)=\arctan(nx)\) converges pointwise on all of \(\mathbb{R}\) to the not-continuous function \[f(x)=\begin{cases} \pi/2, & \quad\text{if }x>0.\\ 0, & \quad\text{if }x=0.\\ -\pi/2, & \quad\text{if }x<0.\ \end{cases}\]

Key: Pointwise convergence does not preserve continuity. We need a better notion.

The following is a generalization of a theorem from Analysis I.

We would like to develop criteria to determine when sequences of complex valued functions converge uniformly. Here is the first, if we already know the sequence converges pointwise:

Checking for uniform convergence (example)

Consider the sequence of functions \(f_{n}(z)=e^{z}+\frac{1}{n}\) and \(g_{n}(z)=e^{z}+\frac{z}{n}\). It is easy to see that both functions converge pointwise to the exponential function \(f(z)=e^{z}\): for every fixed \(z\in\mathbb{C}\) we have \(\lim_{n\to\infty}(e^{z}+\frac{1}{n})=e^{z}=\lim_{n\to\infty}(e^{z}+\frac{z}{n})\).

Notice that \(f_{n}\to f\) uniformly because for every \(z\in\mathbb{C}\) we have \[|f_{n}(z)-f(z)|=\left|\left(e^{z}+\frac{1}{n}\right)-e^{z}\right|=\frac{1}{n};\] so we may take \(s_{n}=1/n\) in Lemma [lem:uniform-convergence-test] part \(1.\).

However \(g_{n}\not\to f\) uniformly - the giveaway is that in writing \[|g_{n}(z)-f(z)|=\left|\left(e^{z}+\frac{z}{n}\right)-e^{z}\right|=\frac{|z|}{n};\] the difference depends on \(|z|\), and \(|z|\) is unbounded in \(\mathbb{C}\). So, we simply notice that upon taking \(z_{n}=n\) that \(|g_{n}(z_{n})-f(z_{n})|=1\) and by Lemma [lem:uniform-convergence-test] part \(2.\) there is no uniform convergence.

However, for \(\rho>0\) the sequence \(g_{n}\) does converge uniformly in any ball \(B_{\rho}(0)\) in the complex plane. For then \(|g_{n}(z)-f(z)|=|z|/n<s_{n}:=\rho/n\).
As mentioned, we are mainly interested in the convergence of infinite series.

Determining the convergence of series via the M-test (example)

Show that \[\sum_{n=1}^{\infty}\frac{|2z|^{3n}}{3^{2n}\:n^{2}}\] converges uniformly on \(\overline{\mathbb{D}}\) and let \(f(z)\) be its limit function. Is \(f(z)\) continuous on \(\overline{\mathbb{D}}\)?

Note that when \(|z|\leq1\) we have \[\frac{|2z|^{3n}}{3^{2n}\:n^{2}}\:\:\leq\:\:\frac{2^{3n}}{3^{2n}\:n^{2}}\:\:=\:\:\left(\frac{8}{9}\right)^{n}\:\left(\frac{1}{n^{2}}\right)\:\:<\:\:\frac{1}{n^{2}}.\] We know \(\sum_{n\in\mathbb{N}}1/n^{2}\) converges, so taking \(M_{n}=1/n^{2}\) the Weierstrass M-test implies the function converges uniformly to some limit function \(f\). Furthermore, since every \(f_{n}\) was continuous, then so is \(f\) (by Theorem [thm:uniform-limit-cont]).
Have we found our ideal definition of convergence? Indeed, we record one other crucial property of uniform convergence from Analysis I:

Everything looks good, but there is a slight issue to think about.

The problem with uniform convergence (an important example)

Let us return to the sequence of functions \(f_{n}(z)=z^{n}\). We saw that it converges pointwise on the (open) unit disc \(\mathbb{D}\) to the function \(f(z)=0\). Is this convergence uniform?

Let’s see if we can find a sequence such that \(|f_{n}(z_{n})-f(z_{n})|=c\) for some \(c>0\). For simplicity, let’s just try to find a sequence of positive real numbers: We want \[c=|f_{n}(z_{n})-f(z_{n})|=|(z_{n})^{n}-0|=(z_{n})^{n},\] so simply take \(z_{n}=c^{1/n}\). We need this sequence to be in \(\mathbb{D}\) so let’s set \(c=1/2\). Then with \(z_{n}=(1/2)^{1/n}\in\mathbb{D}\), by construction we have \[|f_{n}(z_{n})-f(z_{n})|=\left|\left(\left(\frac{1}{2}\right)^{1/n}\right)^{n}-0\right|=\left(\frac{1}{2}\right)^{n/n}=\frac{1}{2}.\] It follows from Lemma [lem:uniform-convergence-test] part \(2.\) that the convergence is not uniform. BUT, notice that the limit function \(f(z)=0\) is trivially continuous on all of \(\mathbb{D}\) - our notion can’t even conclude that the constant function is continuous for a very basic example - we have in some sense been too restrictive!
Key: Uniform convergence is actually too restrictive in terms of preservation of continuity.

Locally uniform but not uniform convergence (example)

Again, consider \(f_{n}(z)=z^{n}\). We know this sequence converges pointwise to \(f=0\) on the (open) unit disc \(\mathbb{D}\), but not uniformly. However, the convergence is locally uniform.

For \(w\in\mathbb{D}\), we can find \(r<1\) such that \(w\in B_{r}(0)\) (this will be our open set \(U\)). Then for all \(z\in B_{r}(0)\), we have \(|z^{n}|<r^{n}\) and \(\lim_{n\to\infty}r^{n}=0\). Hence by Lemma [lem:uniform-convergence-test] part \(1.\) (with \(s_{n}:=r^{n}\)), we have uniform convergence in \(B_{r}(0)\).

Note the ‘counter example’ \((1/2)^{1/n}\) from above fails since \(\lim_{n\to\infty}(1/2)^{1/n}=1\), so at some point the sequence \((1/2)^{1/n}\) leaves the ball \(B_{r}(0)\).

Showing a power series converges (example)

For example, consider the series \(\sum_{n=0}^{\infty}\frac{z^{n}}{n!}\). Applying the ratio test, we have \[L\:=\:\lim_{n\to\infty}\frac{|z^{n+1}|/(n+1)!}{|z^{n}|/n!}\:=\:\lim_{n\to\infty}\frac{|z|}{n+1}\:=\:0\:<\:1,\] for every \(z\in\mathbb{C}\); so \(R=\infty\) and the series converges absolutely in \(\mathbb{C}\). In fact, by Theorem [thm:A-power-series-converges-loc-un] it converges locally uniformly in the whole complex plane to a continuous function. This series looks familiar, and actually agrees with the exponential function \(e^{z}\) we’ve defined. We’ll be able to prove this later in the course.

Differentiation/integration of power series

Now we have an idea of what it means for a power series to converge, and where it converges, we would like to know when a power series represents a holomorphic function. In other words, we want to know when we can differentiate (and integrate) power series. We have the following generalization of a result from Analysis I for real power series.

So, we know the series with the expected formulae for derivative and anti-derivative both converge, but do they genuinely represent the derivative and anti-derivative of the original series?