MR 01 Loesung

Lösung in $\mathbb {C}$

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