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Question Number 206637 by jshfnahdj last updated on 21/Apr/24

Answered by Frix last updated on 21/Apr/24

$$x^y\mathrm{with}x,y\in \mathbb{C}$$$$\mathrm{We}\mathrm{need}x=r{\mathrm{e}}^{\mathrm{i}\theta }\mathrm{and}y=a+b\mathrm{i}\Rightarrow$$$$x^y={\left(r{\mathrm{e}}^{\mathrm{i}\theta }\right)}^{a+b\mathrm{i}}=r^{a+b\mathrm{i}}{\mathrm{e}}^{\mathrm{i}\theta (a+b\mathrm{i})}=$$$$=r^a{r}^{\mathrm{i}b}{\mathrm{e}}^{-b\theta }{\mathrm{e}}^{\mathrm{i}a\theta }={\mathrm{e}}^{a\ln r}{\mathrm{e}}^{\mathrm{i}b\ln r}{\mathrm{e}}^{-b\theta }{\mathrm{e}}^{\mathrm{i}a\theta }=$$$$={\mathrm{e}}^{a\ln r-b\theta }{\mathrm{e}}^{\mathrm{i}(b\ln r+a\theta )}=$$$$={\mathrm{e}}^{a\ln r-b\theta }(\cos (b\ln r+a\theta )+\mathrm{i}\sin (b\ln r+a\theta ))$$$$$$$$1-3\mathrm{i}=\sqrt{10}{\mathrm{e}}^{-{\tan }^{-1}3}$$$$\Rightarrow {(1-3\mathrm{i})}^{-\frac{3}{2}\mathrm{i}}=$$$$={\mathrm{e}}^{-\frac{{3\tan }^{-1}3}{2}}(\cos \frac{3\ln 10}{4}-\mathrm{i}\sin \frac{3\ln 20}{4})\approx$$$$\approx -.0238822-.151706\mathrm{i}$$

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