Let \(U\subset \mathbb C\) be an open set. Define what it means for a function \(f:U\to \mathbb C\) to be complex differentiable at a point \(z_0\in U\).
State the Cauchy-Riemann equations.
Let \(f:U\to \mathbb C\) be the function defined by \[f(z)=f(x+iy)=x\cos(y)+\sinh(iy)\cosh(x).\] Use the Cauchy-Riemann equations to determine the points \(z_0\in\mathbb C\) where \(f\) is complex differentiable.
A function \(f:U\to\mathbb C\) is complex differentiable at \(z_0\in U\) if \[\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}\] exists.
The pair of functions \(u,v:U\to \mathbb R\) satisfy the Cauchy-Riemann equation at \(z_0=x_0+iy_0\) if \[u_x\left( x_0,y_0\right) = v_y\left( x_0,y_0\right),\qquad u_y\left( x_0,y_0\right)=-v_x\left( x_0,y_0\right).\]
We know that a function \(f=u+iv\), with \(u,v:U\to\mathbb R\), is complex
differentiable at \(z_0=x_0+iy_0\in U\)
if and only if \(u\) and \(v\) are differentiable at \(\left( x_0,y_0\right)\) and satisfy the
Cauchy-Riemann equations. In order to identify \(u\) and \(v\) we first need to simplify \(\sinh\left( iy\right)\). We have that \[\sinh\left( iy\right) = \frac{e^{iy}-e^{-iy}}{2}
= i\sin(y)\] and consequently \[f(z) =
x\cos(y)+i\sin(y)\cosh(x)=u(x,y)+iv(x,y)\] where \(u(x,y)=x\cos(y)\) and \(v(x,y)=\sin(y)\cosh(x)\). We have that
\[u_x\left( x,y\right) = \cos(y),\qquad
v_y(x,y)=\cos(y)\cosh(x),\] \[u_y(x,y)=-x\sin(y),\qquad
v_x(x,y)=\sin(y)\sinh(x).\] Using Cauchy-Riemann equation we find
that \[\cos(y)=\cos(y)\cosh(x),\qquad
x\sin(y)=\sin(y)\sinh(x).\] The first equation implies that \(\cos(y)=0\) or \(\cosh(x)=1\), i.e. \(y=\frac{\pi}{2}+n\pi\) with \(n\in\mathbb Z\) or \(x=0\).
When \(y=\frac{\pi}{2}+n\pi\) with
\(n\in\mathbb Z\): we have that the
second equation reads as \[\left( -1\right)^n
x =x\sin\left( \frac{\pi}{2}+n\pi\right)=\sin\left(
\frac{\pi}{2}+n\pi\right)\sinh(x) = \left( -1\right)^n \sinh(x)\]
or \(x=\sinh(x)\). There is only one
solution to this equation, which is \(x=0\).
When \(x=0\): we have that the
second equation reads as \[0=0\cdot
\sin(y)=\sin(y)\sinh(0)=0,\] i.e. the equation always
holds.
In conclusion, Cauchy-Riemann equations hold for any \(\left( x_0,y_0\right)\) such that \(x_0=0\) and as \(u\) and \(v\) are differentiable we conclude that
\(f\) is complex differentiable if and
only if \(z_0\in i\mathbb R\).
On what subset of \(\mathbb C\) is the function \(f(z)=\left( z+i\right)^4-3\) conformal? Justify your response.
Describe the geometric effects of \(f(z)\) on the tangent vectors of the curves passing through the point \(z=1-2i\).
Let \(\gamma:[0,3]\to \mathbb
C\) be the contour given by \[\gamma(t):=\begin{cases}
2t,& \text{if }0\leq t\leq 1,\\
4-2i+2\left( -1+i\right)t,& \text{if }1\leq t\leq 2,\\
2\left( 3-t\right)i,& \text{if }2\leq t\leq 3.
\end{cases}\] (a) Sketch \(\gamma(t)\) in \(\mathbb C\).
(b) Evaluate \(\int_{\gamma}\cos(z)dz\).
We know from class that if a function \(f\) is holomorphic in a domain (i.e. open and connected set) \(U\) then it is conformal at \(z_0\in U\) if and only if \(f'\left( z_0\right)\not=0\). Our function \(f\) is an entire function so we only need to check where the derivative is zero. \[f'(z) = 4\left( z+i\right)^3\] which implies that \(f'(z)\not=0\) if and only if \(z\not=-i\). Consequently, we conclude that \(f\) is conformal on \(\mathbb C\setminus \left\lbrace -i\right\rbrace\).
Given a function \(f\) on a domain \(U\) such that \(f'(z_0)\not=0\), we have that the tangent vectors of the image of a given curve passing by \(z_0\) is the multiplication of the original tangent vector by \(f'(z_0)\) which acts as a stretch by \({\left|f'(z_0)\right|}\) and rotation by \(\mathrm{Arg}\left( f'(z_0)\right)\). In our case \(z_0=1-2i\) gives \[f'\left( 1-2i\right) = 4\left( 1-i\right)^3 = 4 \left( \sqrt{2}e^{-\frac{i\pi}{4}}\right)^3=2^{\frac{7}{2}}e^{-\frac{3i\pi}{4}}.\] We conclude that the geometric effects of \(f(z)\) on the tangent vectors of the curves passing through the point \(z=1-2i\) is a stretch by \(2^{\frac{7}{2}}\) and rotation by \(\frac{3\pi}{4}\) clockwise.
The curve is composed of three straight lines which intersect at the appropriate points: \[\gamma_1:\;\;x(t)=2t,\quad y(t)=0,\quad 0\leq t\leq 1,\] \[\gamma_2:\;\;x(t)=4-2t,\quad y(t)=-2+2t,\quad 1\leq t\leq 2,\] \[\gamma_3:\;\;x(t)=0,\quad y(t)=2(3-t),\quad 2\leq t\leq 3.\]
We know that \(\cos(z)\) is entire and as such is holomorphic in a domain that contains the closed contour \(\gamma\) (\(\mathbb C\)). By the Complex Fundamental Theorem of Calculus we conclude that \[\int_{\gamma}\cos(z)dz=0.\]
Prove that for each \(a\in \mathbb R\), \(a>0\), the series \[\sum_{n=1}^\infty n^{-z}\] converges uniformly on \(\left\lbrace z\in\mathbb C\;:\; \mathrm{Re}(z)>1+a\right\rbrace\) where \(n^{-z}\) is defined using the principal logarithm [You may use without proof that \(\sum_{n=1}^\infty n^{-b}\), \(b\in\mathbb R\), \(b>1\), converges.]
Does the series \(\sum_{n=1}^\infty n^{-z}\) defines a continuous function on \(\left\lbrace z\in\mathbb C\;:\; \mathrm{Re}(z)>1\right\rbrace\)? Justify your response.
Does the series \(\sum_{n=1}^\infty n^{-z}\) converge uniformly on \(\left\lbrace z\in\mathbb C\;:\; \mathrm{Re}(z)\geq 1\right\rbrace\)? Justify your response.
We will aim to use one of Weierstrass M-tests – standard or its local variant. Denote by \(f_n(z)=n^{-z}\). We notice that for any \(z\in \mathbb C\) \[{\left|f_n(z)\right|} = {\left|e^{-z\mathop{\mathrm{Log}}n}\right|} = {\left|e^{-\mathrm{Re}\left( z\right)\log n}e^{-i\mathrm{Im}\left( z\right)\log n}\right|} = e^{-\mathrm{Re}\left( z\right)\log n} = n^{-\mathrm{Re}\left( z\right)}.\] On \(\left\lbrace z\in\mathbb C\;:\; \mathrm{Re}(z)>1+a\right\rbrace\) we have that \[{\left|f_n(z)\right|} \leq n^{-1+a}=M_n.\] Using the given hint we have that \(\sum_{n=1}^{\infty}M_n<\infty\) and consequently, using Weierstrass’ M-test, we conclude that \(\sum_{n=1}^\infty f_n(z)\) converges uniformly on \(\left\lbrace z\in\mathbb C\;:\; \mathrm{Re}(z)>1+a\right\rbrace\).
The idea is similar to the previous result, though we see that we can’t avoid having \(a>0\) in using Weierstrass’ M-test. However, we know that \(\sum_{n=1}^\infty f_n(z)\) converges uniformly on \(\left\lbrace z\in\mathbb C\;:\; \mathrm{Re}(z)>1+a\right\rbrace\) for any \(a>0\). Given \(w\in \left\lbrace z\in\mathbb C\;:\; \mathrm{Re}(z)>1\right\rbrace\) we can find \(a_w>0\) such that \(w\in\left\lbrace z\in\mathbb C\;:\; \mathrm{Re}(z)>1+a_w\right\rbrace\), for example \(a_w = \frac{1+\mathrm{Re}\left( w\right)}{2}\). In other words, for any \(w\in \left\lbrace z\in\mathbb C\;:\; \mathrm{Re}(z)>1\right\rbrace\) there exists an open set \(U_w=\left\lbrace z\in\mathbb C\;:\; \mathrm{Re}(z)>1+a_w\right\rbrace\) that contains \(w\) and on which the series converges uniformly. This means, by definition, that \(\sum_{n=1}^\infty f_n(z)\) converges locally uniformly on \(\left\lbrace z\in\mathbb C\;:\; \mathrm{Re}(z)>1\right\rbrace\). As \(f_n(z)\) are continuous in the domain for any \(n\in\mathbb N\) we conclude from a theorem from class that the resulting function is also continuous.
We notice that when \(z=1\) the series is nothing but the harmonic series \(\sum_{n=1}^\infty n^{-1}\) which doesn’t converge. Consequently the series can’t converge uniformly in \(\left\lbrace z\in\mathbb C\;:\; \mathrm{Re}(z)\geq 1\right\rbrace\) as it doesn’t even converge pointwise there.
Consider the set \(U=\mathbb C\setminus \left\lbrace iy\;:\;y\in\mathbb R,\;y\leq 0\right\rbrace\).
Sketch the set \(U\) in \(\mathbb C\).
Is \(U\) an open set? Justify your response.
Find a biholomorphic map from \(U\) to the open unit disc \(\mathbb{D}=\left\lbrace z\in \mathbb C\;:\; {\left|z\right|}<1\right\rbrace\) and justify why this map is biholomorphic.
The set \(U\) is open. We can
show it by definition, i.e. by showing that for any \(z\in U\) there exists \(\epsilon>0\) such that \(B_\epsilon(z)\subset U\), or by showing
that \(U^c\) is closed.
By definition: Given \(z\in U\)
we have that \(\mathrm{Re}\left(
z\right)\not=0\) or \(z=iy\)
with \(y>0\). In the former case we
choose \(\epsilon=\frac{{\left|\mathrm{Re}(z)\right|}}{2}\)
and find that for any \(w\in
B_{\epsilon}(z)\), \(\mathrm{Re}(w)\not=0\). Indeed, we have
that \[\mathrm{Re}\left( w\right) =
\mathrm{Re}{z}+\mathrm{Re}\left( w-z\right).\] If \(\mathrm{Re}(w)=0\) then \[\mathrm{Re}(z) = \mathrm{Re}\left(
z-w\right)\] which implies that \[{\left|\mathrm{Re}(z)\right|} =
{\left|\mathrm{Re}\left( z-w\right)\right|} \leq {\left|z-w\right|} <
\frac{{\left|\mathrm{Re}(z)\right|}}{2},\] which is impossible.
Consequently, \(B_{\epsilon}(z)\subset
U\).
Consider now the case where \(z=iy\)
with \(y>0\). Then we claim that
\(B_{\frac{y}{2}}(z)\subset U\).
Indeed, for any \(w\in
B_{\frac{y}{2}}(z)\) \[\mathrm{Im}(w)
= \mathrm{Im}(z)+\mathrm{Im}(w-z) = y + \mathrm{Im}(w-z).\] Since
\({\left|\mathrm{Im}(w-z)\right|} \leq
{\left|w-z\right|} < \frac{y}{2}\) we have that \[\mathrm{Im}(w) > y -
\frac{y}{2}=\frac{y}{2}>0,\] which shows that \(w\in U\). As \(w\) was arbitrary we conclude that \(B_{\frac{y}{2}}(z)\subset U\) and with it
the openness of \(U\).
By using \(U^c\): By definition
we find that \[U^c = \mathbb C\setminus U =
\left\lbrace iy\;:\; y\in\mathbb R,\;y\leq 0\right\rbrace.\] To
show that \(U^c\) is closed we will
show that if \(\left\lbrace
z_n\right\rbrace_{n\in\mathbb N}\subset U^c\) converges to a
point \(z\in \mathbb C\), then \(z\in U^c\). Indeed, assuming that \(\left\lbrace z_n\right\rbrace_{n\in\mathbb
N}\) converges to \(z\) implies
that \[\mathrm{Im}\left(
z_n\right)\underset{n\to\infty}{\longrightarrow}\mathrm{Im}(z).\]
Since \(\left\lbrace
z_n\right\rbrace_{n\in\mathbb N}\subset U\) we find that \(\mathrm{Im}\left( z_n\right)\leq 0\) for
all \(n\in\mathbb N\). Consequently,
\(\mathrm{Im}(z)\leq 0\) as the limit
of non-positive sequence. This implies that \(z\in U^c\) and concludes the
proof.
When considering maps to unit spheres powers and Möbius
transformations come to mind. In our case we see that by rotating the
domain by \(\frac{\pi}{2}\) clockwise
we get the domain \[\left\lbrace z\in\mathbb
C\;:\; \mathrm{Re}(z)\leq 0\right\rbrace\] which lends itself
well to powers that use the principal logarithm. This will allow us to
use the principal squared root, \(\sqrt{z}\), which will take the above
domain to the right half plane. At this point we can rotate the domain
by \(\frac{\pi}{2}\) anti-clockwise we
get the upper half plane and use the Cayley map which we know take the
upper half plane to \(\mathbb{D}\).
More formally: Consider the maps \[f_1:\mathbb C\to \mathbb C,\qquad
f_1(z)=e^{-\frac{i\pi}{2}}z\] \[f_2:
\mathbb C\setminus \left\lbrace z\in\mathbb C\;:\; \mathrm{Re}(z)\leq
0\right\rbrace\to \mathbb{H}_R,\qquad f_2(z)=\sqrt{z},\] where
the principal branch of the logarithm was chosen, \[f_3:\mathbb{H}_R\to \mathbb{H},\qquad
f_3(z)=e^{\frac{i\pi}{2}}z,\] \[f_4:\mathbb{H}\to\mathbb{D},\qquad
f_4(z)=\frac{z-i}{z+i}.\] Each of these maps is holomorphic with
a holomorphic inverse in the appropriate domain \[f_1^{-1}(z) = f_3(z),\;\; f_2^{-1}(z)=z^2,\;\;
f_3^{-1}(z)=f_1(z),\;\; f_4^{-1}(z)=i\frac{z+1}{1-z},\] showing
that each map is biholomorphic in the appropriate domain. The desired
map would be \[f(z) = f_4\circ f_3 \circ f_2
\circ f_1 (z).\]