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simple example of composed conformal mapping

\documentclass{article}
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\begin{document}
Let's consider the mapping
$$f\colon \mathbb{C}\to\mathbb{C} \quad \mathrm{with}\quad f(z) = az\!+\!b,$$
where $a$ and $b$ are complex \PMlinkescapetext{constants} and\, $a \neq 0$.

Because\, $f'(z) \equiv a \neq 0$,\, the mapping is conformal in the whole 
$z$-plane.\, Denote\, $\displaystyle a := \varrho e^{i\alpha}$ (where\, $\varrho,\,\alpha \in\mathbb{R}$) and
\begin{align}
z_1 := \varrho z,
\end{align}
\begin{align}
z_2 := e^{i\alpha}z_1,
\end{align}
\begin{align}
w := z_2\!+\!b.
\end{align}
Then the mapping\, $z\mapsto z_1$\, means a dilation in the complex plane, the mapping\, $z_1\mapsto z_2$\, a rotation by the angle $\alpha$ and the mapping\, $z_2\mapsto w$\, a translation determined by the vector from the origin to the point $b$.\, Thus $f$ is composed of these three consecutive mappings which all are conformal.

\begin{figure}
\begin{center}
\includegraphics{simple1}
\end{center}
\caption{The mapping from $z$ to $z_1$}
\end{figure}

\begin{figure}
\begin{center}
\includegraphics{simple2}
\end{center}
\caption{The mapping from $z_1$ to $z_2$}
\end{figure}

\begin{figure}
\begin{center}
\includegraphics{simple3}
\end{center}
\caption{The mapping from $z_2$ to $w$}
\end{figure}

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