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 \(\mathbb{C}-R_{\theta_{1}}\) 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 \(v\in S^{2}-N\), 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 \(P:S^{2}-N\to\mathbb{C}\) 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 \(S^{2}-N\) with \(\mathbb{C}\).

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