UMD Analysis Qualifying Exam/Jan08 Complex

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Problem 2[edit]

Prove there is an entire function  f \!\, so that for any branch g \!\, of  \sqrt{z} \!\,


\sin^2 (g(z))=f(z) \!\,


for all  z \!\, in the domain of definition of  g \!\,

Solution 2[edit]

Key steps

  • \sin^2(\theta)=\frac{1-\cos(2\theta)}{2} \!\,
  • \cos(z)=\sum_{n=1}^\infty \frac{(-1)^nz^{2n}}{(2n)!} \!\,
  • ratio test

Problem 4[edit]

Let H\!\, be the domain \{ z: -\frac{\pi}{2} < \Re(z) < \frac{\pi}{2}, \Im(z)>0 \}\!\,. Prove that g=\sin(z)\!\, is a 1:1 conformal mapping of H\!\, onto a domain D\!\,. What is D\!\,?

Solution 4[edit]

Showing G 1:1 conformal mapping[edit]

First note that



\begin{align}
(1) \quad |g^{\prime}(z)| &= |\sin^{\prime}(z)|\\
                &= |\cos(z)| \\
                &= \left|\frac{e^{-iz}+e^{iz}}{2}\right| \\
                &\geq \frac{1}{2} (|e^{-ix}||e^{y}|-|e^{ix}||e^{-y}| )  \\
                &\geq \frac{1}{2}(e^y-e^{-y}) \\
                &> 0 \quad \mbox{since }y>0
\end{align}


Also, applying a trigonometric identity, we have for all  z_1, z_2 \in H \!\,,


 (2) \quad \sin z_1 - \sin z_2 = 2 \sin \left(\frac{z_1-z_2}{2}\right)\cos\left(\frac{z_1+z_2}{2}\right) \!\,


Hence if \sin z_1=\sin z_2 \!\,, then


\sin \left(\frac{z_1-z_2}{2}\right)=0 \!\,


or


\cos \left(\frac{z_1+z_2}{2}\right)=0 \!\,


The latter cannot happen in  H \!\, since  |g^{\prime}(z)|=|\cos(z)|>0 \!\, so


\sin \left(\frac{z_1-z_2}{2}\right)=0 \!\,


i.e.


 z_1 = z_2 \!\,


Note that the zeros of \sin(z)=\sin(x+iy) \!\, occur at x=k \pi,  k \in \mathbb{Z} \!\,. Similary the zeros of \cos(z)=\cos(x+iy) \!\, occur at x=\frac{\pi}{2}+k\pi, k \in \mathbb{Z} \!\,.


Therefore from (1) \!\, and (2) \!\,, g\!\, is a 1:1\!\, conformal mapping.

Finding the domain D[edit]

To find  D \!\,, we only need to consider the image of the boundaries.


Consider the right hand boundary,  C_3=\{z=x+iy | x=\frac{\pi}{2}, y >0 \} \!\,



\begin{align}
g(C_3) &= g\left(\frac{\pi}{2}+ yi\right)\\
       &= \sin\left(\frac{\pi}{2}+ yi\right) \\
       &= \frac{e^{i(\frac{\pi}{2}+yi)}-e^{-i(\frac{\pi}{2}+yi})}{2i}\\
       &= \frac{e^{i\frac{\pi}{2}}e^{-y}-e^{-i\frac{\pi}{2}}e^{y}}{2i} \\
       &= \frac{e^{i\frac{\pi}{2}}e^{-y}+e^{i\frac{\pi}{2}}e^{y}}{2i} \\
       &= \frac{e^{i\frac{\pi}{2}}(e^{-y}+e^y)}{2i} \\
       &= \frac{i(e^{-y}+e^y)}{2i} \\
       &= \frac{e^{-y}+e^y}{2} \\
\end{align}


Since  y>0 \!\,,


g(C_3)=(1,\infty) \!\,


Now, consider the left hand boundary  C_2=\{z=x+iy | x=-\frac{\pi}{2}, y >0 \} \!\,.



\begin{align}
g(C_2) &= g\left(-\frac{\pi}{2}+ yi\right)\\
       &= \sin\left(-\frac{\pi}{2}+ yi\right) \\
       &= \frac{e^{i(-\frac{\pi}{2}+yi)}-e^{-i(-\frac{\pi}{2}+yi})}{2i}\\
       &= \frac{e^{-i\frac{\pi}{2}}e^{-y}-e^{i\frac{\pi}{2}}e^{y}}{2i} \\
       &= \frac{-e^{i\frac{\pi}{2}}e^{-y}-e^{i\frac{\pi}{2}}e^{y}}{2i} \\
       &= \frac{e^{i\frac{\pi}{2}}(-e^{-y}-e^y)}{2i} \\
       &= -\frac{i(e^{-y}+e^y)}{2i} \\
       &= -\frac{e^{-y}+e^y}{2} \\
\end{align}


Since  y>0 \!\,,


g(C_2)=(-\infty,-1) \!\,


Now consider the bottom boundary  C_1=\{z=x+iy | -\frac{\pi}{2}<= x <= \frac{\pi}{2}, y=0 \} \!\,.



\begin{align}
g(C_1) &= g(x)\\
       &= \sin(x) 
\end{align}


Since  -\frac{\pi}{2}<= x <= \frac{\pi}{2} \!\,,


g(C_2)=[-1,1] \!\,


Hence, the boundary of H \!\, maps to the real line. Using the test point  z=i \!\,, we find



\begin{align}
g(i) &= \sin(i) \\
     &= \frac{e^{i(i)}-e^{-i(i)}}{2i}\\
     &= \frac{e^{-1}-e^{1}}{2i}\\
     &= \frac{-i(e^{-1}-e^{1})}{2}\\
     &= \frac{i(e^{1}-e^{-1})}{2}\\
     &= \frac{i}{2}(e-\frac{1}{e}) \\
     &\in \mbox{Upper Half Plane}
\end{align}


We then conclude D=g(H)=\mbox{Upper Half Plane} \!\,

Problem 6[edit]

Suppose that for a sequence  a_n \in R \!\, and any  z, \Im (z) >0 \!\,, the series


 h(z) = \sum_{n=1}^\infty a_n \sin (nz) \!\,


is convergent. Show that  h \!\, is analytic on  \{ \Im(z)> 0 \} \!\, and has analytic continuation to  C \!\,


Solution 6[edit]

Summation a_n Convergent[edit]

We want to show that \sum_{n=1}^\infty a_n \!\, is convergent. Assume for the sake of contradiction that \sum_{n=1}^\infty a_n \!\, is divergent i.e.


\sum_{n=1}^\infty a_n = \infty\!\,


Since h(z) \!\, is convergent in the upper half plane, choose  z=i \!\, as a testing point.



\begin{align}
h(i)   &= \sum_{n=1}^\infty a_n \sin(ni) \\
       &= \sum_{n=1}^\infty a_n \frac{e^{-n}-e^n}{2i} \\
       &= \sum_{n=1}^\infty a_n \frac{-i(e^{-n}-e^n)}{2} \\
       &= \sum_{n=1}^\infty a_n \frac{i(e^n-e^{-n})}{2} 
\end{align}


Since h(i) \!\, converges in the upper half plane, so does its imaginary part and real part.


 \Im(h(i)) = \frac12 \sum_{n=1}^\infty a_n 
\underbrace{(e^n-e^{-n})}_{E_n}
\!\,

The sequence \{E_n\} \!\, is increasing (E_1 < E_2 < \ldots < E_n < E_{n+1} \!\,) since e^{n+1} > e^n \!\, and e^{-n} > e^{-(n+1)} \!\, e.g. the gap between e^n \!\, and e^{-n} \!\, is grows as n \!\, grows. Hence,



\begin{align}
\Im(h(i)) &\geq \frac12 (e^1 - e^{-1}) \underbrace{\sum_{n=1}^\infty a_n}_{= \infty} \\
\\
\\
\Im(h(i)) &\geq \infty 
\end{align}


This contradicts that h(i) \!\, is convergent on the upper half plane.

Show that h is analytic[edit]

In order to prove that h \!\, is analytic, let us cite the following theorem


Theorem Let {h_n}\!\, be a sequence of holomorphic functions on an open set U\!\,.   Assume that for each compact subset K\!\, of U\!\, the sequence converges uniformly on K\!\,, and let the limit function be h\!\,.   Then h\!\, is holomorphic.

Proof See Theorem 1.1 in Chapter V, Complex Analysis Fourth Edition, Serge Lang.


Now, define h_n(z)= \sum_{k=1}^{n} a_k \sin(k z)\!\,.   Let K\!\, be a compact set of U=\{\Im{z}>0\}\!\,.   Since  \sin(kz) \!\, is continuous