Komplexes Potenzieren/Hoch 3/Reelle Koordinaten/Harmonisch/Aufgabe/Lösung

a) Es ist

{}z^{3}=(x+{\mathrm {i} }y)^{3}=x^{3}+3{\mathrm {i} }x^{2}y-3xy^{2}-{\mathrm {i} }y^{3}=x^{3}-3xy^{2}+{\left(3x^{2}y-y^{3}\right)}{\mathrm {i} }\,,

die Komponentenfunktionen sind also ${}g(x,y)=x^{3}-3xy^{2}$ und ${}h(x,y)=3x^{2}y-y^{3}$ .

b) Es ist

{}{\begin{aligned}{\frac {\partial ^{2}g}{\partial ^{2}x}}+{\frac {\partial ^{2}g}{\partial ^{2}y}}&={\frac {\partial ^{2}{\left(x^{3}-3xy^{2}\right)}}{\partial ^{2}x}}+{\frac {\partial ^{2}{\left(x^{3}-3xy^{2}\right)}}{\partial ^{2}y}}\\&=6x-6x\\&=0\end{aligned}}

und

{}{\begin{aligned}{\frac {\partial ^{2}h}{\partial ^{2}x}}+{\frac {\partial ^{2}h}{\partial ^{2}y}}&={\frac {\partial ^{2}{\left(3x^{2}y-y^{3}\right)}}{\partial ^{2}x}}+{\frac {\partial ^{2}{\left(3x^{2}y-y^{3}\right)}}{\partial ^{2}y}}\\&=6y-6y\\&=0.\end{aligned}}