MR 01 Loesung

Lösung in $\mathbb {C}$

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

$\forall z\in \mathbb {C} \setminus \{0\}$