\documentclass[a4paper,12pt]{article}
\begin{document}
\parindent=0pt
QUESTION In each of the following case state whether or not the
function $f$ of the complex variable $z=x+iy$ is (i) continuous,
(ii) differentiable at $z=0$ and give a brief justification for
each answer.
\begin{description}
\item[(a)]
$f(z)=|z|$,
\item[(b)]
$f(z)=xy$
\item[(c)]
$f(z)=\frac{x^2-y^2}{x^2+y^2}$ for $z\neq0$ and $f(0)=1.$
\item[(d)]
$f(z)=\frac{\sinh z}{z},\ z\neq 0$ and $f(0)=1$.
\end{description}
ANSWER
\begin{description}
\item[(a)]
Continuous at 0 as $\lim_{z\to0}|z|=0(=|0|)$. Not differentiable
at 0 as $\lim_{h\to 0}\frac{|h|-|0|}{h}=\frac{|h|}{h}$ and this
means that the limit does not exist as it depends on the path we
use to reach zero.
\item[(b)]
Continuous at 0(same reasons as in (a).) Is is differentiable at 0
as the Cauchy-Riemann equations hold at 0, and the partial
derivatives exist in a neighbourhood of 0 and are continuous
there. (Theorem 3.3)
\item[(c)]
Change to polar coordinates, $x=r\cos \theta, y=r\sin\theta$. Then
$f(z)=\cos 2\theta$. Hence not continuous and hence not
differentiable at 0.
\item[(d)]
It is easy to see that L'h\^{o}pital's rule applies in complex
variables as the same proof as in real variables holds and this
shows that $f$ is continuous at ). It is differentiable at 0 for
reasons that will be apparent later in the course. However, you
could just use the definition and then we have to examine
$\lim_{z\to0}\frac{f(z)-f(0)}{z}$. This is equal to $\lim_{z\to
0}\frac{\sinh z-z}{z^2}$, and you can find this (and prove it
exists) by using L'H\^{o}pital twice. You also need to show that
the partial derivatives of $f$ exist in a neighbourhood of 0 and
are continuous there.
\end{description}
\end{document}