UMD Analysis Qualifying Exam/Jan08 Complex

Problem 2

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

Key steps

• $\sin ^{2}(\theta )={\frac {1-\cos(2\theta )}{2}}\!\,$

• $\cos(z)=\sum _{n=1}^{\infty }{\frac {(-1)^{n}z^{2n}}{(2n)!}}\!\,$

• ratio test

Problem 4

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

Showing G 1:1 conformal mapping

First note that

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

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

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

Consider the right hand boundary, $C_{3}=\{z=x+iy|x={\frac {\pi }{2}},y>0\}\!\,$

${\begin{aligned}g(C_{3})&=g\left({\frac {\pi }{2}}+yi\right)\\&=\sin \left({\frac {\pi }{2}}+yi\right)\\&={\frac {e^{i({\frac {\pi }{2}}+yi)}-e^{-i({\frac {\pi }{2}}+yi})}{2i}}\\&={\frac {e^{i{\frac {\pi }{2}}}e^{-y}-e^{-i{\frac {\pi }{2}}}e^{y}}{2i}}\\&={\frac {e^{i{\frac {\pi }{2}}}e^{-y}+e^{i{\frac {\pi }{2}}}e^{y}}{2i}}\\&={\frac {e^{i{\frac {\pi }{2}}}(e^{-y}+e^{y})}{2i}}\\&={\frac {i(e^{-y}+e^{y})}{2i}}\\&={\frac {e^{-y}+e^{y}}{2}}\\\end{aligned}}$

Since $y>0\!\,$,

$g(C_{3})=(1,\infty )\!\,$

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

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

Since $y>0\!\,$,

$g(C_{2})=(-\infty ,-1)\!\,$

Now consider the bottom boundary $C_{1}=\{z=x+iy|-{\frac {\pi }{2}}<=x<={\frac {\pi }{2}},y=0\}\!\,$.

${\begin{aligned}g(C_{1})&=g(x)\\&=\sin(x)\end{aligned}}$

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

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

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

Problem 6

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

Summation a_n Convergent

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.

${\begin{aligned}h(i)&=\sum _{n=1}^{\infty }a_{n}\sin(ni)\\&=\sum _{n=1}^{\infty }a_{n}{\frac {e^{-n}-e^{n}}{2i}}\\&=\sum _{n=1}^{\infty }a_{n}{\frac {-i(e^{-n}-e^{n})}{2}}\\&=\sum _{n=1}^{\infty }a_{n}{\frac {i(e^{n}-e^{-n})}{2}}\end{aligned}}$

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

$\Im (h(i))={\frac {1}{2}}\sum _{n=1}^{\infty }a_{n}\underbrace {(e^{n}-e^{-n})} _{E_{n}}\!\,$

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,

${\begin{aligned}\Im (h(i))&\geq {\frac {1}{2}}(e^{1}-e^{-1})\underbrace {\sum _{n=1}^{\infty }a_{n}} _{=\infty }\\\\\\\Im (h(i))&\geq \infty \end{aligned}}$

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

Show that h is analytic

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 $h_{n}(z)=\sum _{k=1}^{n}a_{k}\sin(kz)\!\,$.   Let $K\!\,$ be a compact set of $U=\{\Im {z}>0\}\!\,$.   Since $\sin(kz)\!\,$ is continuous