# simple example of composed conformal mapping

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 and  $a\neq 0$.

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

 $\displaystyle z_{1}:=\varrho z,$ (1)
 $\displaystyle z_{2}:=e^{i\alpha}z_{1},$ (2)
 $\displaystyle w:=z_{2}\!+\!b.$ (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.

Title simple example of composed conformal mapping SimpleExampleOfComposedConformalMapping 2013-03-22 16:47:25 2013-03-22 16:47:25 pahio (2872) pahio (2872) 8 pahio (2872) Example msc 30E20 msc 53A30