Complex Analysis II, Michaelmas 2024. Problem Class 4

Denoting by \(M_C\) the Cayley map, we know that \(\left( M_C\right)^{-1}=M_{C^{-1}}\) takes \(\mathbb D\) to \(\mathbb{H}\). Consequently, the Möbius map \(M=M_{C^{-1}}\circ f\) takes \(\mathbb{H}\) to itself. By the H2H theorem we know that we can find \(S\in SL_2\left( \mathbb R\right)\) such that \[M_{C^{-1}}\circ f = M_S\] and consequently (using the fact that \(M_C \circ M_{C^{-1}}=M_{Id}=Id\)) we have that \[f=M_C\circ M_S = M_{CS}.\] Writing \(f=M_T\) for some \(T\in GL_2\left( \mathbb C\right)\) and using the fact that \(C=\begin{pmatrix} 1 & -i \\ 1 & i \end{pmatrix}\) we see that if \(S=\begin{pmatrix} a & b\\ c& d \end{pmatrix}\) with \(a,b,c,d\in \mathbb R\) and \(ad-bc=1\) we can choose \(T\) to be \[T=\begin{pmatrix} a-ic & b-id \\ a+ic & b+id \end{pmatrix}\] with \(a,b,c,d\in \mathbb R\) and \(ad-bc=1\). As \(f\) takes \(1+i\) to \(0\) and \(1\) to \(-i\) we have that \[\left( a-ic\right)\left( 1+i\right) + b-id=0\] and \[a-ic + b-id = -i\left( a+ic + b+id\right)\] we find that \[-i\left( a-ic\right) = -i\left( a+ic + b+id\right)\] which implies that \(-2ic = b+id\) and \(a=\frac{d-b}{2}\). Lastly, we see that \[1=ad-bc = \frac{d^2-bd}{2} + \frac{db-ib^2}{2}=\frac{d^2-ib^2}{2}.\] We conclude that \(d=\pm \sqrt{2}\), \(b=0\) and consequently \(a=\frac{\pm\sqrt{2}}{2}\) and \(c=\mp \frac{\sqrt{2}}{2}\). Plugging this back yields \[T=\pm \frac{\sqrt{2}}{2}\begin{pmatrix} 1+i & -2i \\ 1-i & 2i \end{pmatrix}\] which, due to the scaling invariance, shows that \[f(z) = \frac{\left( 1+i\right)z -2i}{\left( 1-i\right)z +2i}.\]

We have seen that the Cayley map \(M_C\) takes the first quadrant \(\Omega_1 = \left\lbrace z\in \mathbb C\;:\; \mathop{\text{Im}}(z)>0, \mathop{\text{Re}}(z)>0\right\rbrace\) to the lower half of \(\mathbb{D}\). We conclude that \(g(z)=-M_C(z)\) takes \(\Omega_1\) to the upper half of \(\mathbb{D}\) and consequently \(g^{-1}(z) = M_{C^{-1}}(-z)\) takes the upper half of \(\mathbb{D}\) to \(\Omega_1\). Continuing on the above, we notice that \(z^2\) is a biholomrphic map that takes \(\Omega_1\) to \(\mathbb{H}\) and by using the Cayley map again we see that \[f(z) = M_C\left( \left( g^{-1}(-z)\right)^2\right) = M_C\left( M_{C^{-1}}(-z)^2\right)\] takes the upper half of \(\mathbb{D}\) to \(\mathbb{D}\) in a bioholomorphic way. Using the fact that \(M_C(z)=\frac{z-i}{z+i}\) and \(M_{C^{-1}}(z)=\frac{iz+i}{-z+1}\) we find that \[f(z)=\frac{\left( z-1\right)^2+i\left( z+1\right)^2}{\left( z-1\right)^2-i\left( z+1\right)^2}.\]

We see that we are asked to map half a plane to a circle so the Cayley map pops to mind. Consider the map \(f_1(z) = z-\frac{i}{2}\). We have that \(f_1\) takes \(\mathcal{R}\) to the lower half plane and as such \(-f_1\) takes \(\mathcal{R}\) to \(\mathbb{H}\). We conclude that \[f_2(z) = M_C\left( -f_1(z)\right)\] takes \(\mathcal{R}\) to \(\mathbb{D}\). Next we notice that \(f_3(z) = \frac{z-1}{2}\) takes \(\mathcal{R}'\) to \(\mathbb{D}\) as well and is invertible with \(f^{-1}_3(z) = 2z+1\). As all the above maps are biholomorphic we find that \[f(z) = 2M_C\left( -z+\frac{i}{2}\right)+1\] takes \(\mathcal{R}\) to \(\mathcal{R}'\) in a biholomorphic way. Using the fact that \(M_C(z)=\frac{z-i}{z+i}\) we find that \[f(z)=\frac{6z-i}{2z-3i}.\]

(i) For all \(z\in\mathbb C^\ast\) we have that \[\lim_{n\to\infty}f_n(z)=\lim_{n\to\infty}\frac{1}{nz} = 0\] and as such \(\left\lbrace f_n(z)\right\rbrace_{n\in\mathbb N}\) converges pointwise to \(f(z)=0\). On \(A\) we find that \[{\left|f_n(z)-f(z)\right|} = \frac{1}{n{\left|z\right|}} \leq \frac{1}{n\rho}=s_n.\] We know that \(\left\lbrace s_n\right\rbrace_{n\in\mathbb N}\) is a positive sequence that is independent of \(z\) and converges to zero. We conclude, according to a lemma from class, that \(\left\lbrace f_n(z)\right\rbrace_{n\in\mathbb N}\) converges to zero uniformly on \(A\).
(ii) The convergence on \(\mathbb C^\ast\) will not be uniform as we can get as close as we want to \(z=0\) where the sequence of functions is unbounded. Indeed, choosing \(z_n=\frac{1}{n}\) we see that \(\left\lbrace z_n\right\rbrace_{n\in\mathbb N}\subset \mathbb C^\ast\) and \[{\left|f_n(z_n)-f(z_n)\right|} = 1.\] We conclude, according to a lemma from class, that \(\left\lbrace f_n(z)\right\rbrace_{n\in\mathbb N}\) does not converge to zero uniformly on \(A\).

We know that for any (fixed) \(z\in \mathbb C\) \[\lim_{n\to\infty}\frac{z}{n}=0\] (not uniformly!). As \(\sin(z)\) is a continuous function we conclude that for any (fixed) \(z\in\mathbb C\) \[\lim_{n\to \infty}\sin\left( \frac{z}{n}\right)=\sin\left( 0\right)=0,\] showing that \(f_n(z)\) converges to \(f(z)=0\) pointwise on \(\mathbb C\).
To show the uniform convergence we recall that \[\sin(z)=\sin(x)\cosh(y)+i\cos(x)\sinh(y)\] and as such \[f_n(z)=\sin\left( \frac{z}{n}\right) = \sin\left( \frac{x}{n}\right)\cosh\left( \frac{y}{n}\right)+i\cos\left( \frac{x}{n}\right)\sinh\left( \frac{y}{n}\right).\] We conclude that \[{\left|f_n(z)-f(z)\right|} = \sqrt{\sin^2\left( \frac{x}{n}\right)\cosh^2\left( \frac{y}{n}\right)+\cos^2\left( \frac{x}{n}\right)\sinh^2\left( \frac{y}{n}\right)}.\] At this point it is important to (yet again) remind ourselves that \(\sin(z)\) is not bounded in general due to the appearance of \(\sinh\) and \(\cosh\). We do find that \[{\left|f_n(z)-f(z)\right|} \leq \sqrt{\sin^2\left( \frac{x}{n}\right)\cosh^2\left( \frac{y}{n}\right)+\sinh^2\left( \frac{y}{n}\right)}\] (where we have kept the terms that we know will go to zero) and using the facts that \(\sin(-x)=-\sin(x)\), \(\sinh(-x)=\sinh(x)\), and \(\cosh(-x)=\cosh(x)\) we find that \[{\left|f_n(z)-f(z)\right|} \leq \sqrt{\sin^2\left( {\left|\frac{x}{n}\right|}\right)\cosh^2\left( {\left|\frac{y}{n}\right|}\right)+\sinh^2\left( {\left|\frac{y}{n}\right|}\right)}\] Using the fact that on \(A\) \[\max\left\lbrace {\left|x\right|},{\left|y\right|}\right\rbrace \leq {\left|z\right|} \leq \rho\] together with the facts that \(\sinh\) and \(\cosh\) are increasing on \([0,\infty)\) and \({\left|sin(x)\right|} \leq {\left|x\right|}\) for all \(x\in\mathbb R\) we find that on \(A\) \[{\left|f_n(z)-f(z)\right|} \leq \sqrt{\left( \frac{\rho}{n}\right)^2\cosh^2\left( \frac{\rho}{n}\right)+\sinh^2\left( \frac{\rho}{n}\right)}=s_n.\] We know that \(\left\lbrace s_n\right\rbrace_{n\in\mathbb N}\) is a positive sequence that is independent of \(z\) and converges to zero. We conclude, according to a lemma from class, that \(\left\lbrace f_n(z)\right\rbrace_{n\in\mathbb N}\) converges to zero uniformly on \(A\).
The lack of uniform convergence on \(\mathbb C\) can be seen by considering the sequence \(z_n=n\). Indeed \[{\left|f_n(z_n)-f(z_n)\right|} = {\left|\sin(1)\right|}\underset{n\to\infty}{\not\longrightarrow}0.\] We can do “worse” and even get unboundedness. For instance, choosing \(z_n=in^2\) (to bring out the \(\sinh\) and \(\cosh\) which only depend on \(\mathop{\text{Im}}(z)\)) gives us \[{\left|f_n(z_n)-f(z_n)\right|} = {\left|\sinh(n)\right|}=\sinh(n)\] which goes to infinity as \(n\) goes to infinity.

For any \(z\in A\) we have that \[{\left|e^{nz}\right|} = e^{n\mathop{\text{Re}}\left( z\right)} \leq e^{-n} = \left( e^{-1}\right)^n=M_n.\] As \(\sum_{n=0}^\infty M_n <\infty\) (a geometric series with \(q=e^{-1}<1\)) we conclude by Weierstrass’ M-test that the series converges uniformly on \(A\).
The reason we will have issues with the convergence on \(B\) is the fact that the real part of \(z\) can be zero. Indeed, \(z=0\) is in \(B\) and \[\sum_{n=0}^{\infty} e^{n\cdot 0}= \sum_{n=0}^\infty 1 =\infty\] so the series doesn’t even converge on \(B\).
It is worth to mention that the series does converge locally uniformly on \(B^0 = \{z \in \mathbb C: \: \mathop{\text{Re}}(z)< 0\}\). Indeed, given any \(z\in B^0\) we consider the open set \(U_z = \left\lbrace w\in\mathbb C\;:\; \mathop{\text{Re}}(w) < \frac{\mathop{\text{Re}}(z)}{2}\right\rbrace\). Clearly \(z\in U_z\) and for any \(w\in U_z\) we have that \[{\left|e^{nw}\right|} = e^{n\mathop{\text{Re}}\left( w\right)} \leq e^{\frac{n\mathop{\text{Re}}(z)}{2}} = \left( e^{-\frac{{\left|\mathop{\text{Re}}(z)\right|}}{2}}\right)^n=M_n\left( U_z\right).\] As \(\sum_{n=0}^\infty M_n\left( U_z\right) <\infty\) we conclude the result by using Weierstrass’ local M-test.

We recall that the reverse triangle inequality states that \[{\left|z-w\right|} \geq {\left|{\left|z\right|}-{\left|w\right|}\right|}.\] Using the above inequality, we conclude that if \({\left|z\right|}<R<1\) then \[{\left|1+z^n\right|} = {\left|1-(-z^n)\right|} \geq {\left|1-{\left|-z^n\right|}\right|} = {\left|1-{\left|z\right|}^n\right|} = 1-{\left|z\right|}^n \geq 1-R^n,\] where we used the fact that \({\left|z\right|}^n < R^n <1\) as \(R<1\).
For any \(z\in A\) we have that \[{\left|\frac{z^n}{1+z^n}\right|} =\frac{{\left|z^n\right|}}{{\left|1+z^n\right|}} \leq \frac{R^n}{1-R^n}=M_n.\] Since \(M_n>0\) and \[\frac{M_{n+1}}{M_n} = \frac{1-R^{n}}{1-R^{n+1}}\;R\underset{n\to\infty}{\longrightarrow}\frac{1-0}{1-0}\;R=R<1\] we conclude that \(\sum_{n=0}^\infty M_n<\infty\). Using Weierstrass’ M-test, we find that \(\sum_{n=1}^{\infty}\frac{z^{n}}{1+z^n}\) converges uniformly on \(A\). Since \(\frac{z^n}{1+z^n}\) are continuous on \(A\) (as we do not include the boundary \({\left|z\right|}=1\) where the denominator is unbounded) we conclude that the limit function of the series is continuous by the same theorem.