Wegintegral/C/z^2dz/Weg/Beispiel

Wir betrachten die holomorphe Differentialform ${}z^{2}dz$ und den Weg

$\gamma \colon [0,1]\longrightarrow {\mathbb {C} },\,t\longmapsto t^{2}-t{\mathrm {i} }.$

Es ist ${}\gamma '(t)=2t-{\mathrm {i} }$ und somit ist

${}{\begin{aligned}\int _{\gamma }z^{2}dz&=\int _{0}^{1}(\gamma (t))^{2}\cdot \gamma '(t)dt\\&=\int _{0}^{1}{\left(t^{2}-t{\mathrm {i} }\right)}^{2}{\left(2t-{\mathrm {i} }\right)}dt\\&=\int _{0}^{1}{\left(t^{4}-t^{2}-2{\mathrm {i} }t^{3}\right)}{\left(2t-{\mathrm {i} }\right)}dt\\&=\int _{0}^{1}{\left(2t^{5}-2t^{3}-2t^{3}+{\left(-t^{4}+t^{2}-4t^{4}\right)}{\mathrm {i} }\right)}dt\\&=\int _{0}^{1}{\left(2t^{5}-4t^{3}+{\left(-5t^{4}+t^{2}\right)}{\mathrm {i} }\right)}dt\\&=\left({\frac {1}{3}}t^{6}-t^{4}\right)|_{0}^{1}+{\mathrm {i} }\left(-t^{5}+{\frac {1}{3}}t^{3}\right)|_{0}^{1}\\&=-{\frac {2}{3}}-{\frac {2}{3}}{\mathrm {i} }.\end{aligned}}$