simple example of composed conformal mapping

Let’s consider the mapping

$$f:ℂ\to ℂ\mathit{\hspace{1em}}\mathrm{with}\mathit{\hspace{1em}}f(z)=az+b,$$

where $a$ and $b$ are complex and  $a\ne 0$.

Because  $f^{\prime }(z)\equiv a\ne 0$,  the mapping is conformal in the whole$z$-plane.  Denote  $a:=\varrho e^{i\alpha }$ (where  $\varrho ,\alpha \in ℝ$) and

$z_1:=\varrho z,$(1)

$z_2:=e^{i\alpha }{z}_1,$(2)

$w:=z_2+b.$(3)

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.

Figure 1: The mapping from $z$ to $z_1$

Figure 2: The mapping from $z_1$ to $z_2$

Figure 3: The mapping from $z_2$ to $w$