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\noindent{\it Math 6720}
\hfill{\bf Appl. Complex Var., Asymptc Mthds}\hfill
{\it Alexander Balk}


\bigskip\bigskip

\noindent{\bf HW5}\hfill{\bf\large NAME:
.....................................................................}
\hfill {\bf due 2/17/2016}

\bigskip
\begin{enumerate}

%\item Suppose we would like to consider the {\it analytic continuation} of
%  some function $f(z)$ given by a power series
%\begin{eqnarray*}
%f(z)=\sum_{n=0}^\infty a_n (z-b_0)^n,\quad\quad
%\mbox{ where }\quad\quad z\in D_0: |z-b_0|<R_0
%\end{eqnarray*}
%(this is the largest disk that the series converges). To build 
%analytic continuation, we choose a point $b_1\in D_0$ and compute
%$f(b_1), f'(b_1), f''(b_1), \ldots$ by means of the given power
%series; then we set up the new power series
%\begin{eqnarray*}
%\sum_{n=0}^\infty \frac{f^{(n)}(b_1)}{n!} (z-b_1)^n,\quad\quad
%\mbox{ where }\quad\quad z\in D_1: |z-b_1|<R_1
%\end{eqnarray*}
%(this is the largest disk that the series centered at $b_1$
%converges).
%\begin{enumerate}
%\item Show that $R_1\geq R_0 - |b_1-b_0|$. 
%\item If $R_1> R_0 - |b_1-b_0|$, we have analytically continued function
%$f(z)$ beyond the disk $D_0$. Show that if $R_1= R_0 - |b_1-b_0|$, the only
%common point of the two circles $\partial D_0$ and $\partial D_1$ 
%is a singularity of $f(z)$.
%\end{enumerate}
%\end{enumerate}

%\item
%Expand
%$$ f(z)=\frac{z}{(z-2)(z+i)}$$
%in a Laurent series (centered at $z_0=0$) in the domains
%\begin{enumerate}
%\item $|z|<1$,
%\item $1<|z|<2$,
%\item $|z|>2$.
%\end{enumerate}


\item
Let $z_0$ be an isolated singularity of an analytic function $f(z)$. Near
$z_0$, the function $f(z)$ can be represented by its Laurent series (L)
\begin{eqnarray*}
f(z)=\sum_{n=-\infty}^{\infty} a_n (z-z_0)^n.
\end{eqnarray*}
Prove:
\begin{enumerate}
\item If (L) has no negative powers, then $z_0$ is a removable singularity.


\item If $z_0$ is a removable singularity, then (L) has no negative
  powers.

{\footnotesize{\it Suggestion:} You should actually prove a little stronger statement:\\
If $f(z)$ is bounded in some neighborhood
around $z_0$, then (L) has no negative powers. 

{\it Idea of the proof:} Consider the formula for the (L) coefficients given as the integral over a small
circle $|z-z_0|=R\rightarrow 0$; \\
estimate for $n<0$. This is similar to the proof of the Liouville theorem.} 
  
\bigskip

\item If (L) has a finite number of negative powers, then $z_0$ is a pole.
\item If $z_0$ is a pole, then (L) has a finite number of negative powers.

\smallskip
{\footnotesize{\it Suggestion:} 
Expand $g(z)=1/f(z)$ in the Taylor series around $z_0$.} 

\bigskip

\item If (L) has infinitely many negative powers, then $z_0$ is an
  essential singularity.
\item If $z_0$ is an essential singularity, then (L) has infinitely many 
  negative powers.

\smallskip
{\footnotesize{\it Suggestion:} 
The last two statements follow from the previous ones.}

\end{enumerate}

\newpage\item
Prove Kasorati-Sokhotsky-Weierstrass Theorem:

If an analytic function $f(z)$ has an essential singularity at $z_0$, \\
then $f(z)$ comes arbitrarily close to {\it any} complex value in every neighbourhood of $z_0$;\\
in other words,\\
$\forall$ $\epsilon > 0,\;\;\forall \delta >0$, and $\forall$ complex number $w$, 
$\exists$ a complex number $z$ such that \\$|z-z_0￼| < \delta$ and $|f(z)-w| < \epsilon$.

\smallskip

{\footnotesize{\it Suggestion:} 
Assume on the contrary and consider bounded function $g(z)=[f(z)-w]^{-1}$; expand it in the Taylor series.\\
Express $f(z)$ via $g(z)$ 
and arrive at contradiction that $f(z)$ either has a pole or removable singularity at $z_0$.}

\newpage\item The (single-valued) function $f(z)=\sqrt{z(z-1)(z-4)}$ is defined by
\begin{eqnarray*}
\mbox{branch cuts }\quad z=-i t,\quad z=1+i t, \quad z=4-i t
\quad\quad (0\le t <\infty)
\end{eqnarray*}
and value $f(2)=-2i$.

\bigskip

Evaluate
\begin{enumerate}
\item$f(-3)$,
\item$f(1/2)$,
\item$f(5)$.
\end{enumerate}

\newpage\item 
The (single-valued) function 
\begin{eqnarray*}
f(z)=\frac{(z-16)^{1/3}(z-20)^{1/2}}{z-1}
\end{eqnarray*}
is defined by
\begin{eqnarray*}
\mbox{branch cuts }\quad z=16+i t,\quad z=20-i t
\quad\quad (0\le t <\infty)
\end{eqnarray*}
and value $f(24)=\frac{4}{23}$.

\bigskip

Evaluate $\oint_C f(z) dz$ over the circle $C: \; |z-2|=2$.

\newpage\item 
We are given multivalued analytic function
 $$f(z)=[(z-1)(z-2)]^{1/3}.$$
What branch cuts would you make to separate single-valued
branches?\\
Would the cut from $z=1$ to $z=2$ along the real axis work?

%\newpage\item
%Let a function $f(z)$ be some branch of $z^{1/n}$ ($n$ is an integer). 
%Show that $f'(z)=\frac{f(z)}{n\; z}$.\\
%In other words, the function and its derivative are determined by {\it the same branch} of $z^{1/n}$.

\newpage\item
The (single-valued) function $f(z)=\sqrt{z(z-1)}$ is defined by
the branch cut $z=t\;(0\le t\le 1)$ and the condition that 
$f$ is positive for real $z>1$.

\begin{enumerate}
\item Find the first three non-zero terms of its Laurent series (centered
at $z_0=0$) in the domain $|z|>1$.
\item Where does the series converge?
\end{enumerate}


\newpage\item
Evaluate the integral
\begin{eqnarray*}
\int_{C}z^{-1/3}dz
\end{eqnarray*}
over the path $C$ traversing from $z=1+i$ to $z=1-i$ along the
following composite curve $C=C_1+C_2+C_3$
\begin{eqnarray*}
&&C_1:\quad \mbox{hyperbola } 2y^2-x^2=1 
          \mbox{ from } z=1+i \mbox{ to } z=-2+i\sqrt{5/2},\\
&&C_2:\quad \mbox{segment } x=-2 
          \mbox{ from } z=-2+i\sqrt{5/2} \mbox{ to } z=-2-i\sqrt{5/2},\\
&&C_3:\quad \mbox{hyperbola } 2y^2-x^2=1 
          \mbox{ from } z=-2-i\sqrt{5/2} \mbox{ to } z=1-i.
\end{eqnarray*}

Here $z^{-1/3}$ denotes the branch defined by the branch cut along the positive $x$-axis 
and the condition that it is real when $z$ is negative.

\newpage\item Show that
\begin{eqnarray*}
  \int_0^{2\pi}\frac{d\phi}{(p+q\cos\phi)^2}=
\frac{2\pi p}{(p^2-q^2)^{3/2}}
\end{eqnarray*}
(with constant parameters $p>q>0$).

\newpage\item Show that 
\begin{eqnarray*}
 \int_0^\infty\frac{x^m}{x^n+1} \; dx =
\frac{\pi}{n\;\sin\frac{\pi(m+1)}{n} }\;
\end{eqnarray*}
with integer parameters $m$ and $n$, $\;\;n-2\ge m\ge 0$.

{\footnotesize{\it Suggestion:} 
Notice that the expression under the integral sign is multiplied by a
constant factor when $x$ is changed to $x e^{i 2\pi/n}$. Consider
residue integration in the sector between 
$z=r$ and $z=r e^{i 2\pi/n}$ ($0\le r<\infty$). 
The integrand has only one pole in this sector.}

\newpage\item {\bf Jordan's lemma} to neglect integrals
\begin{eqnarray*}
  I(R)=\int_{C_R}f(z)\; e^{i a z} \; dz\quad\quad (a>0)
\end{eqnarray*}
over ``large'' semi-circle $C_R: \; z=Re^{i\phi}\;\;
(0\le\phi\le\pi)$, $\quad R\rightarrow\infty$.

\bigskip
{\bf Prove:}\hspace*{1.3cm}
If on $C_R$, $\quad|f(z)|<F(R)\rightarrow0\quad$  
[for some function $F(R)$], 
\hspace*{1cm} then $I(R)\rightarrow 0$. 



\end{enumerate}
\end{document}


\newpage\item
Let $z_0$ be an isolated singularity of analytic function $f(z)$. Near
$z_0$, the function $f(z)$ can be represented by its Laurent series (L)
\begin{eqnarray*}
f(z)=\sum_{n=-\infty}^{\infty} a_n (z-z_0)^n.
\end{eqnarray*}
Prove:
\begin{enumerate}
\item If L has no negative powers, then $z_0$ is a removable singularity.
\item If $z_0$ is a removable singularity, then L has no negative
  powers.
\item A statement stronger than (b): If $f(z)$ is bounded in some neighborhood
around $z_0$ (w/o $z_0$ itself), then L  has no negative powers. 

{\footnotesize{\it Suggestion:} 
Consider the formula for the L coefficients given as the integral over a small
circle $|z-z_0|=R\rightarrow 0$; estimate for $n<0$. This is similar
to your proof of the Liouville theorem.} 

\bigskip

\item If L has finite number of negative powers, then $z_0$ is a pole.
\item If $z_0$ is a pole, then L has finite number of negative powers.\\
{\footnotesize{\it Suggestion:} 
Expand $g(z)=1/f(z)$ in the Taylor series around $z_0$.} 

\bigskip

\item If L has infinitely many negative powers, then $z_0$ is an
  essential singularity.
\item If $z_0$ is an essential singularity, then L has infinitely many 
  negative powers.\\
{\footnotesize{\it Suggestion:} 
The last two statements follow from the previous ones.}

\end{enumerate}

\newpage\item
Describe singularities (their locations and types) for the
  function
$$ f(z)=\frac{(z-1)^2\;\cos^2 (z/2)}{(z+1)^3\; z(z-\pi)}.$$

\newpage\item Consider the multi-valued function 
$$f(z)=\frac{\sqrt{z}-1}{z-1}.$$ 
To separate $f(z)$ into
single-valued branches, make a brunch cut along the negative real
semi-axis. 
What isolated singularities (their locations and types)
do these branches have?

\newpage\item Consider the multi-valued function 
$$f(z)=(1+z)^{1/z}.$$ 
To separate $f(z)$ into
single-valued branches, make a brunch cut $z=-t\; (1\le t<\infty)$. 
What isolated singularities (their locations and types)
do these branches have?


\newpage\item
The function $f(z)$ is analytic in the entire complex plane except at
$z=i/2$, where it has a simple pole, and at $z=2$, where it has a
double pole. It is known that
\begin{eqnarray*}
\oint_{|z|=1}f(z)dz &=& 2\pi i\,,\\
\oint_{|z|=3}f(z)dz &=& 0\,, \\
\oint_{|z|=3}f(z)(z-2)dz &=& 0\,,
\end{eqnarray*}
and $f(z)$ is bounded at infinity 
(i.e. $\exists M>0, \;\exists R>0: |z|>R\;\Rightarrow\; |f(z)|<M$).

\medskip

Find $f(z)$ (unique up to an arbitrary additive constant).

{\footnotesize{\it Suggestion:} 
Find an examle $f_0(z)$ of such function and use the Liouville theorem
to show that $f(z)-f_0(z)$ is constant.}

\newpage\item 
Prove that if all singularities of an analytic function $f(z)$ 
in the extended complex plane are poles, then $f(z)$ is a rational function 
(i.e. the ratio of two polynomials).

{\footnotesize{\it Suggestion:} 
First show that $f(z)$ can have only finite number of poles. Then for each
pole at finite $z$ take the negative power part of the corresponding Laurent
expansion; for $z=\infty$, take the positive power part; add all of
them to get a rational function $R(z)$. Finally, apply the Liouville theorem
to show that $f(z)-R(z)$ is a constant.}


\end{enumerate}



\end{document}



Suppose we would like to approximate the branches of the multi-valued function 
$$f(z)=(1+z)^{1/z}$$ for small $z\rightarrow 0$ by three-term expansions
 (i.e. the ones with 3 non-zero terms) in powers of $z$ . 
Find the three-term approximation for that branch 
of this multi-valued function which takes real values for all positive $z$.
 
Is it possible to find the three-term expansions approximating the
 other branches?





\newpage\item 
