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Question Number 156186 by DavidSmath last updated on 09/Oct/21

Commented by DavidSmath last updated on 09/Oct/21

$$\mathrm{show}\mathrm{clear}\mathrm{workings}...$$

Commented by MJS_new last updated on 09/Oct/21

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{see}\mathrm{how}\mathrm{there}\mathrm{could}\mathrm{be}\mathrm{more}\mathrm{than}$$$$\mathrm{one}\mathrm{value}.{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{same}\mathrm{as}\mathrm{in}\mathrm{the}\mathrm{following}$$$$\mathrm{example}:$$$$x=2^2\Rightarrow x=4$$$$x={(-2)}^2\Rightarrow x=4$$$$\mathrm{but}$$$$x^2=4\Rightarrow x=-2\vee x=2$$$$$$$$\mathrm{if}\mathrm{we}\mathrm{have}{z}_1^{z_2}\mathrm{we}\mathrm{need}{z}_1=r{\mathrm{e}}^{\mathrm{i}\theta }\mathrm{with}r\in \mathbb{R}\mathrm{and}$$$$0⩽\theta <2\pi \mathrm{and}{z}_2=a+b\mathrm{i}$$$$\Rightarrow$$$${\left(r{\mathrm{e}}^{\mathrm{i}\theta }\right)}^{a+b\mathrm{i}}=r^{a+b\mathrm{i}}{\mathrm{e}}^{\mathrm{i}\theta (a+b\mathrm{i})}=r^a{r}^{b\mathrm{i}}{\mathrm{e}}^{\mathrm{i}a\theta }{\mathrm{e}}^{-b\theta }=$$$$=r^a{\mathrm{e}}^{-b\theta }{\mathrm{e}}^{\mathrm{i}(a\theta +b\ln r)}\mathrm{which}\mathrm{is}\mathrm{unique}.$$

Answered by mr W last updated on 09/Oct/21

$$z={(2i-1)}^{i+1}={\left[\sqrt{5}{e}^{(\pi -{\tan }^{-1}2)i}\right]}^{i+1}$$$$=\sqrt{5}{e}^{-\pi +{\tan }^{-1}2}{\left[\sqrt{5}{e}^{(\pi -{\tan }^{-1}2)}\right]}^i$$$$=\sqrt{5}{e}^{-\pi +{\tan }^{-1}2}{e}^{(\pi -{\tan }^{-1}2+\ln \sqrt{5})i}$$$${it}^{\prime }sprincipalvalueis$$$$Arg\left(z\right)=\pi -{\tan }^{-1}2+\ln \sqrt{5}\approx 2.839$$

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