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\newcommand{\ds}{\displaystyle}
\newcommand{\p}{\partial}
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\begin{document}

{\bf Question}

Explain, without proofs, the relationship between
differentiability of a function of a complex variable and the
Cauchy-Riemann equations.

Prove that the real and imaginary parts of a differentiable
function satisfy Laplace's equation.

Let $u(x,y)=\sin x\sinh y$.

Show that $u$ is a harmonic function.  Find a function $v$ such
that $f=u+iv$ is a differentiable function of $z=x+iy$.

Write down an expression for the derivative of $f$ in terms of
$z$.


\vspace{0.25in}

{\bf Answer}

Let $f(z)=u(x,y)+iv(x,y)$ where $z=x+iy$.  The function $f$ is
differentiable at $z_0$ if $\ds\frac{f(z_0+h)-f(z_0)}{h}$ tends to
a limit as $h\to0$.

The function $f$ satisfies the Cauchy-Riemann equations at $z_0$
if

$\ds\frac{\p u}{\p x}=\frac{\p v}{\p y}$ and $\ds\frac{\p u}{\p
y}=-\frac{\p v}{\p x}$ at $z_0$.

If $f$ is differentiable then it satisfies the Cauchy-Riemann
equations.  The converse is false in general, but we have a
partial converse.  If $f$ satisfies the Cauchy-Riemann equations
at $z_0=x_0+iy_0$, and if the partial derivatives exist in a
neighbourhood of $(x_0,y_0)$ and are continuous at $(x_0,y_0)$
then $f$ is differentiable at $z_0$.

Now if $f$ is differentiable then $f$ is analytic, so partial
derivatives of all orders exist and are continuous

$\begin{array}{lcl}\ds\frac{\p u}{\p x}=\frac{\p v}{\p y}
&amp;\Rightarrow&amp; \ds\frac{\p^2u}{\p x^2}=\frac{\p^2v}{\p x\p y}\\
\ds\frac{\p u}{\p y}=-\frac{\p v}{\p x} &amp;\Rightarrow&amp;
\ds\frac{\p^2u}{\p y^2}=-\frac{\p^2v}{\p y\p x}\end{array}$

Mixed partial derivatives are equal, so adding gives

$\ds\frac{\p^2u}{\p x^2}+\frac{\p^2u}{\p y^2}=0$

Also

$\begin{array}{lcl}\ds\frac{\p u}{\p x}=\frac{\p v}{\p y}
&amp;\Rightarrow&amp; \ds\frac{\p^2u}{\p y\p x}=\frac{\p^2v}{\p y^2}\\
\ds\frac{\p u}{\p y}=-\frac{\p v}{\p x} &amp;\Rightarrow&amp;
\ds\frac{\p^2u}{\p x\p y}=-\frac{\p^2v}{\p x^2}\end{array}$

subtracting and equating mixed derivatives gives

$\ds\frac{\p^2v}{\p x^2}+\frac{\p^2v}{\p y^2}=0$

Now $u=\sin x\sinh y$

$\begin{array}{lcl}\ds\frac{\p u}{\p x}=\cos x\sinh y &amp; &amp;
\ds\frac{\p u}{\p y}=\sin x\cosh y\\ \ds\frac{\p^2u}{\p x^2}=-\sin
x\sinh y &amp; &amp; \ds\frac{\p^2v}{\p y^2}=\sin x\sinh y\end{array}$

So $\ds\frac{\p^2u}{\p x^2}+\frac{\p^2u}{\p y^2}=0$ i.e. $u$ is
harmonic.

$\ds\frac{\p u}{\p x}=\cos x\sinh y=\frac{\p v}{\p y}$

so $v=\cos x\cosh y+\phi(x)$

$-\frac{\p u}{\p y}=-\sin x\cosh y=\frac{\p v}{\p x}$

so $v=\cos x\cosh y+\psi(y)$

so $\phi(x)=\psi(y)$=constant

Thus $v=\cos x\cosh y+c$

\begin{eqnarray*} f &amp;=&amp; \sin x\sinh y+i\cos x\cosh y\\ &amp;=&amp; \sin
x(-i\sin iy)+i\cos x\cos iy\\ &amp;=&amp; i(\cos x\cos iy-\sin x\sin iy)\\
&amp;=&amp; i\cos z\end{eqnarray*}

So $\ds\frac{df}{dz}=-i\sin z$

${}$

Or $\ds\frac{df}{dz}=\frac{\p u}{\p x}+i\frac{\p v}{\p x}$

$=\cos x\sinh y-i\sin x\cosh y$

$=-i(\sin x\cos iy+\cos x\sin iy)$

$=-i\sin z$

\end{document}
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