Question Number 216485 by Tawa11 last updated on 08/Feb/25
$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}\:\:\mathrm{in}:\:\:\:\mathrm{i}^{\mathrm{x}} \:\:=\:\:\mathrm{2} \\ $$
Answered by issac last updated on 09/Feb/25
$${x}\centerdot\mathrm{ln}\left({i}\right)=\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$\mathrm{ln}\left({i}\right)=\frac{\pi}{\mathrm{2}}{i} \\ $$$$\therefore\:{x}=−\frac{\mathrm{2}}{\pi}{i}\centerdot\mathrm{ln}\left(\mathrm{2}\right) \\ $$
Commented by Tawa11 last updated on 09/Feb/25
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$
Answered by Ghisom last updated on 09/Feb/25
$${x}\in\mathbb{C} \\ $$$${x}={a}+{b}\mathrm{i},\:\:{a},\:{b}\:\in\mathbb{R} \\ $$$$\mathrm{i}=\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} \\ $$$$\mathrm{i}^{{x}} =\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}\left({a}+{b}\mathrm{i}\right)} =\mathrm{e}^{−\frac{{b}\pi}{\mathrm{2}}+\mathrm{i}\frac{{a}\pi}{\mathrm{2}}} =\frac{\mathrm{1}}{\mathrm{e}^{\frac{{b}\pi}{\mathrm{2}}} }\mathrm{e}^{\mathrm{i}\frac{{a}\pi}{\mathrm{2}}} =\mathrm{2} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{e}^{\frac{{b}\pi}{\mathrm{2}}} }=\mathrm{2}\:\wedge\:\frac{{a}\pi}{\mathrm{2}}=\mathrm{2}{n}\pi \\ $$$$\Rightarrow\:\mathrm{b}=−\frac{\mathrm{2ln}\:\mathrm{2}}{\pi}\:\wedge\:{a}=\mathrm{4}{n} \\ $$$${x}=\mathrm{4}{n}−\frac{\mathrm{2ln}\:\mathrm{2}}{\pi}\mathrm{i};\:{n}\in\mathbb{Z} \\ $$
Commented by Tawa11 last updated on 09/Feb/25
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$