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Question Number 192580 by Frix last updated on 21/May/23

$$\mathrm{This}\mathrm{might}\mathrm{be}\mathrm{helpful}:$$$$\mathrm{How}\mathrm{to}\mathrm{easily}\mathrm{calculate}{z}_1^{z_2}\mathrm{with}{z}_1,z_2\in \mathbb{C}:$$$$\mathrm{Transform}{z}_1=r{\mathrm{e}}^{\mathrm{i}\theta }\mathrm{and}{z}_2=a+b\mathrm{i}\Rightarrow$$$$z_1^{z_2}={\left(r{\mathrm{e}}^{\mathrm{i}\theta }\right)}^{a+b\mathrm{i}}$$$$z_1^{z_2}=r^{a+b\mathrm{i}}{\mathrm{e}}^{-b\theta +\mathrm{i}a\theta }$$$$z_1^{z_2}=r^a{\mathrm{e}}^{-b\theta }{r}^{b\mathrm{i}}{\mathrm{e}}^{\mathrm{i}a\theta }$$$$z_1^{z_2}=\frac{r^a}{{\mathrm{e}}^{b\theta }}{\mathrm{e}}^{\mathrm{i}(a\theta +b\ln r)}$$

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