MR 01 Loesung

Lösung in $\mathbb {C}$

$f_{1}(z)$	=	${\frac {\left|z\right|^{2}}{z}}$	=	${\bar {z}}$
$f_{2}(z)$	=	${\frac {i\left|z\right|^{2}}{z}}$	=	$\Re <>\Im (z)$
$f_{3}(z)$	=	${\frac {z^{2}+\left|z\right|^{2}}{2z}}$	=	$\Re (z)$
$f_{4}(z)$	=	${\frac {z^{2}-\left|z\right|^{2}}{2z}}$	=	$\Im (z)$