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{\bf Question}

Let $C$ be a circle in the $z$-plane having its centre on the real
axis, and suppose further that it passes through $z=1$ with $z=-1$
as an interior point. Let $C$ be transformed by the mapping
$w=f(z)=\ds\frac{1}{2}\left(z+\ds\frac{1}{z}\right)$. Show that
this map is not conformal at $z=1$. Expand $f(z)$ locally in the
neighbourhood of $z=1$ and deduce that the angles between lines
which meet at $z=1$ are doubled. Hence sketch the image of $C$ in
the neighbourhood if $w=f(1)$. By picking a few other points and
working out their $w$-images, draw the whole of the transformed
$C$. How does the image of $C$ change if its centre is moved to
the upper half plane, but $C$ still passes through $z=1$?

\medskip

{\bf Answer}

(z)

PICTURE \vspace{2in}

$w=f(z)=\ds\frac{1}{2}\left(z+\ds\frac{1}{z}\right) \Rightarrow
\ds\frac{dw}{dz}=\ds\frac{1}{2}\left(1-\ds\frac{1}{z^2}\right)$

so not conformal at $z=\pm 1$ (or 0).

Hence $z=+1$ is a problem point. We expect the angles of $C$ not
to be conserved near the image point $w=f(1)=1$.

\newpage
We carry out a local analysis:

\begin{eqnarray*} f(z) & = &
f(1)+f'(1)(z-1)+f''(1)\ds\frac{(z-1)^2}{2}+\cdots\\ & = &
1+\ds\frac{f''(1)(z-1)^2}{2}+\cdots\\ \Rightarrow w-1 & = &
\ds\frac{(z-1)^2}{2}+\cdots \end{eqnarray*}

Thus if we set $(z-1)=re^{i\theta}$

$\Rightarrow w-1 \approx \ds\frac{r^2}{2}e^{2i\theta}$, i.e., the
angles are \un{doubled}.

Now what is the angle of $C$ at $z=1$?

Well it's formed by the tangents of $C$ at 1:

PICTURE \vspace{1.5in}

So, the regular $C$ at $z=1$ is transformed to a $2\pi$ cusp at
$w=1$. All other points of $C$ transform \un{smoothly} since 0 and
$-1$ are not on the curve.

By symmetry the new curve is symmetric about the $Re(w)$ axis:

PICTURE \vspace{1.5in}

If the centre of $C$ is removed, but we still pass through $z=1$,
the cusp remains, but is bent to a more aerofoil shape.

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