Lineare Gleichung/(-4+3i)z ist (-2+5i)/Betrag/Aufgabe/Lösung

Das Inverse von ${\displaystyle {}-4+3{\mathrm {i} }}$ ist

${\displaystyle {}{\frac {1}{25}}{\left(-4-3{\mathrm {i} }\right)}=-{\frac {4}{25}}-{\frac {3}{25}}{\mathrm {i} }\,.}$

Daher ist

${\displaystyle {}{\begin{aligned}z&={\left(-{\frac {4}{25}}-{\frac {3}{25}}{\mathrm {i} }\right)}{\left(-2+5{\mathrm {i} }\right)}\\&={\frac {8}{25}}+{\frac {15}{25}}+{\left(-{\frac {20}{25}}+{\frac {6}{25}}\right)}{\mathrm {i} }\\&={\frac {23}{25}}-{\frac {14}{25}}{\mathrm {i} }.\end{aligned}}}$

Dabei ist

${\displaystyle {}\vert {z}\vert ={\frac {\sqrt {23^{2}+14^{2}}}{25}}={\frac {\sqrt {529+196}}{25}}={\frac {\sqrt {725}}{25}}={\frac {\sqrt {29}}{5}}\,.}$