(Assignment sheet 4 problem 17) Let \(f(z)\) be a holomorphic function. Prove the
following variants of the Zero derivative
theorem, which says that, if any one of the following
conditions hold on a domain \(D\) then
\(f(z)\) is constant on \(D\).
(i) \(f(z)\) is a real number for all
\(z \in D\).
(ii) the real part of \(f(z)\) is
constant on \(D\).
(iii) the modulus of \(f(z)\) is
constant on \(D\).
(i) In this setting we have that \(f(x+iy)=u(x,y)+iv(x,y)\) and \(v\equiv 0\). As such, using the
Cauchy-Riemann equations we find that \[u_x=0=u_y\] which implies that \(u\equiv const\) and consequently \(f\equiv const\).
(ii) In this setting we have that \(f(x+iy)=u(x,y)+iv(x,y)\) and \(u\equiv const_1\). As such, using the
Cauchy-Riemann equations we find that \[v_x=0=v_y\] which implies that \(v\equiv const_2\) and as such \(f\equiv const\).
(iii) In this setting we have that \(f(x+iy)=u(x,y)+iv(x,y)\) and \[u^2+v^2=c\] for some constant \(c\geq 0\). If \(c=0\) then we must have \(u=v\equiv 0\) and the result is concluded.
We assume, thus, that \(c>0\) and
notice that it implies that we can’t have a point \(x_0+iy_0 \in D\) such that \(u(x_0,y_0)=v(x_0,y_0)=0\) (else \(c=0\)). Moving towards using the
Cauchy-Riemann equations we differentiate the above with respect to
\(x\) and \(y\) to find that \[uu_x+vv_x=0,\] and \[uu_y+vv_y=0.\] The above can be rewriten
as the system \[\begin{pmatrix}
u_x & v_x \\
u_y & v_y
\end{pmatrix} \begin{pmatrix}
u \\y
\end{pmatrix} = \begin{pmatrix}
0 \\0
\end{pmatrix}.\] Since \(u_x=v_y\) and \(u_y=-v_x\) we find that the matrix \(A(x,y) =\begin{pmatrix}
u_x & v_x \\
u_y & v_y
\end{pmatrix}\) satisfies \[\det
A(x,y) = u_xv_y -v_xu_y = u_x^2+u_y^2.\] As our system has a
non-trivial solution for every \(x,y\)
(since \(c>0\)) we must have that
\[0=\det A(x,y)=u_x^2+u_y^2\] for all
\(x,y\) in the domain. Consequently
\(u_x=u_y=0\) in the domain and as such
\(u\equiv const\). We can use point
(ii) or redo the proof for \(v\) to
conclude the result.
(Assignment sheet 5 problem 5) In what subset of the complex plane is \(\sinh z\) conformal? For every point \(z_0\) at which the function is not conformal, give an example of two paths (lines) through \(z_0\) such that the angle (or the orientation of the angle) between them is not preserved by \(f(z)\) at \(z_0\).
Since \(f\) is an entire function
the points where \(f\) is conformal are
the points where \(f'(z)\not=0\).
We have that \(0=f'(z)=\cosh(z)=\frac{e^z+e^{-z}}{2}\)
if and only if \(e^z=-e^{-z}\) or
alternatively \(e^{2z}=-1=e^{i\pi}\).
We conclude that \[2z = i\pi +2\pi i
k\] or \[z=i\left( \frac{\pi}{2}+\pi
k\right)\] for some \(k\in \mathbb
Z\). We conclude that the points where \(f\) is not conformal are points in the set
\[S=\left\lbrace z\in \mathbb C\;:\;
z=i\left( \frac{\pi}{2}+\pi k\right)\text{ for some }k\in\mathbb
Z\right\rbrace.\] For a fixed \(k\) consider the orthogonal lines \[\ell_1:\;\;x=t,\; y=\frac{\pi}{2}+\pi k,\quad
t\in \mathbb R,\] \[\ell_2:\;\; x=0,\;
y=s,\quad s\in \mathbb R.\] The lines meet at \(i\left( \frac{\pi}{2}+\pi k\right)\) which
corresponds to \(t=0\) and \(s=\frac{\pi}{2}+\pi k\). Using the fact
that \[f(x+iy)=\sinh(x+iy)=\sinh(x)\cos(y)+i\cosh(x)\sin(y)\]
The image of these curves is given by \[f\left( \ell_1\right):\;\; u=0, v=\left(
-1\right)^k\cosh(t)\] and \[f\left(
\ell_2\right):\;\; u=0 , v=\sin(s)\] \(f\left( \ell_1\right)\) is the set \(\left\lbrace w\in \mathbb C\;:\;
\mathop{\text{Im}}{w}\geq 1\right\rbrace\) when \(k\) is even and \(\left\lbrace w\in \mathbb C\;:\;
\mathop{\text{Im}}{w}\leq -1\right\rbrace\) when \(k\) is odd, while \(f\left( \ell_2\right)\) is the set \(\left\lbrace w\in \mathbb C\;:\; -1\leq
\mathop{\text{Im}}{w}\leq 1\right\rbrace\).
The point of intersection is \[f\left(
i\left( \frac{\pi}{2}+\pi k\right)\right) = \sinh\left( i\left(
\frac{\pi}{2}+\pi k\right)\right)=-i\sin\left( i\cdot i\left(
\frac{\pi}{2}+\pi k\right)\right) = i\left( -1\right)^k\] and the
angle between the images of the curves is \(\pi\).
(Assignment sheet 6 problem 5) Is there a Möbius transformation which maps the sides of the triangle with vertices at \(-1,i\) and \(1\) to the sides of an equilateral triangle (all sides of equal length)? Either give an example of such a Möbius transformation, or explain why it is not possible.
There is no such Möbius transformation. One way to show this is to try and construct such transformation using the cross ration but a more geometric solution would be to recall that Möbius transformations are conformal at any point they are defined in (on \(\mathbb C\)) and that they take a boundary of a set to the boundary of its image - in our case the given triangle needs to go to the equilateral one. Consequently, the angle between the images of the lines connecting \(-1\) and \(i\) and \(1\) and \(i\) must be perpendicular to one another which is impossible.
(Assignment sheet 6 problem 8) If \(\alpha\) and \(\beta\) are the two fixed points of a Möbius transformation \(f(z)\), show that for all \(z\neq \alpha,\beta\) and \(f(z)\neq\infty\), we have \[\frac{f(z)-\alpha}{f(z)-\beta} =K\frac{z-\alpha}{z-\beta},\] where \(K\) is a constant.
We know that \(f(\alpha)=\alpha\) and \(f(\beta)=\beta\). Let \(z_0\not= \alpha,\beta\) be such that \(w_0=f(z_0)\not=\infty\) (also \(w_0\not= \alpha,\beta\) since Möbius maps are injective). Using the invariance of Möbius transformations under the cross ratio we conclude that for any \(z\in \mathbb C\) \[\left( f(z),w_0;\alpha,\beta\right) = \left( z,z_0;\alpha,\beta\right).\] Equivalently \[\frac{f(z)-\alpha}{f(z)-\beta} \;\frac{w_0-\beta}{w_0-\alpha} = \frac{z-\alpha}{z-\beta}\;\frac{z_0-\beta}{z_0-\alpha}.\] Denoting by \[K=\frac{z_0-\beta}{z_0-\alpha}\; \frac{w_0-\alpha}{w_0-\beta}\] gives us the desired result.
(Assignment sheet 6 problem 9) Find the Möbius transformation taking the ordered set of points \(\{-i,-1,i \}\) to the ordered set of points \(\{-i,0,i\}\). What is the image of the unit disc under this map? Which point is sent to \(\infty\)?
Using the cross ratio we find that \[\frac{f(z)-0}{f(z)-i}\; \frac{-i-i}{-i-0} = \frac{z-(-1)}{z-i}\; \frac{-i-i}{-i-(-1)}\] which implies that \[\frac{f(z)}{f(z)-i} = -\frac{i}{1-i}\;\frac{z+1}{z-i}\] or \[f(z)\left( 1-i\right)\left( z-i\right) = -\left( f(z)-i\right)i\left( z+1\right).\] This can be rewritten as \[f(z)\left[ \left( 1-i\right)\left( z-i\right) + i\left( z+1\right)\right]= -\left( z+1\right)\] or equivalently \[f(z)=-\frac{z+1}{z -1}.\] The image of the unit circle is a circline. As the point which is sent to infinity by \(f\), \(z=1\), is on the circle we conclude that the image of the unit circle is a line. Since the boundary of the unit disc is sent to the boundary of its image under \(f\) and since \(f(0)=1\) we conclude that the image of the unit disc under the map is the right hand-side plane \(\mathbb{H}_R\).