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Question Number 33988 by abdo imad last updated on 28/Apr/18

$$givethealgebricformof{(1+i)}^i.$$

Answered by MJS last updated on 28/Apr/18

$${(1+\mathrm{i})}^{\mathrm{i}}={\left(\sqrt{2}{\mathrm{e}}^{\mathrm{i}\frac{\pi }{4}}\right)}^{\mathrm{i}}={\left({\mathrm{e}}^{\frac{\ln 2}{2}}{\mathrm{e}}^{\mathrm{i}\frac{\pi }{4}}\right)}^{\mathrm{i}}={\left({\mathrm{e}}^{\frac{\ln 2}{2}+\mathrm{i}\frac{\pi }{4}}\right)}^{\mathrm{i}}=={\left(\sqrt{2}{\mathrm{e}}^{\mathrm{i}\frac{\pi }{4}}\right)}^{\mathrm{i}}$$$$={\mathrm{e}}^{-\frac{\pi }{4}+\mathrm{i}\frac{\ln 2}{2}}={\mathrm{e}}^{-\frac{\pi }{4}}{\mathrm{e}}^{\mathrm{i}\frac{\ln 2}{2}}={\mathrm{e}}^{-\frac{\pi }{4}}{\left(\sqrt{2}\right)}^{\mathrm{i}}$$$$\sqrt{2}{\mathrm{e}}^{\mathrm{i}\frac{\pi }{4}}=\sqrt{2}(\cos \frac{\pi }{4}+\mathrm{i}\sin \frac{\pi }{4})$$$${\mathrm{e}}^{-\frac{\pi }{4}}{\mathrm{e}}^{\mathrm{i}\frac{\ln 2}{2}}={\mathrm{e}}^{-\frac{\pi }{4}}(\cos \frac{\ln 2}{2}+\mathrm{i}\sin \frac{\ln 2}{2})$$

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