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Question Number 216485 by Tawa11 last updated on 08/Feb/25

$$\mathrm{Solve}\mathrm{for}\mathrm{x}\mathrm{in}:{\mathrm{i}}^{\mathrm{x}}=2$$

Answered by issac last updated on 09/Feb/25

$$x\cdot \ln \left(i\right)=\ln \left(2\right)$$$$\ln \left(i\right)=\frac{\pi }{2}i$$$$\therefore x=-\frac{2}{\pi }i\cdot \ln \left(2\right)$$

Commented by Tawa11 last updated on 09/Feb/25

$$\mathrm{Thanks}\mathrm{sir}$$

Answered by Ghisom last updated on 09/Feb/25

$$x\in \mathbb{C}$$$$x=a+b\mathrm{i},a,b\in \mathbb{R}$$$$\mathrm{i}={\mathrm{e}}^{\mathrm{i}\frac{\pi }{2}}$$$${\mathrm{i}}^x={\mathrm{e}}^{\mathrm{i}\frac{\pi }{2}(a+b\mathrm{i})}={\mathrm{e}}^{-\frac{b\pi }{2}+\mathrm{i}\frac{a\pi }{2}}=\frac{1}{{\mathrm{e}}^{\frac{b\pi }{2}}}{\mathrm{e}}^{\mathrm{i}\frac{a\pi }{2}}=2$$$$\Rightarrow \frac{1}{{\mathrm{e}}^{\frac{b\pi }{2}}}=2\wedge \frac{a\pi }{2}=2n\pi$$$$\Rightarrow \mathrm{b}=-\frac{2\ln 2}{\pi }\wedge a=4n$$$$x=4n-\frac{2\ln 2}{\pi }\mathrm{i};n\in \mathbb{Z}$$

Commented by Tawa11 last updated on 09/Feb/25

$$\mathrm{Thanks}\mathrm{sir}$$

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