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Question Number 25152 by tawa tawa last updated on 05/Dec/17

$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{real}\mathrm{part}\mathrm{and}\mathrm{imaginary}\mathrm{part}\mathrm{of}\mathrm{the}\mathrm{complex}\mathrm{number}:\mathrm{z}={(1+\mathrm{i})}^{\mathrm{i}}$$

Commented by tawa tawa last updated on 05/Dec/17

$$\mathrm{please}\mathrm{help}.$$

Answered by jota+ last updated on 05/Dec/17

$$z={(1+i)}^i$$$$lnz=iln(1+i)=i\left[ln\left(\sqrt{2}{e}^{i\pi /4}\right)\right]$$$$=i[ln\sqrt{2}+i\frac{\pi }{4}]=(-\frac{\pi }{4}+iln\sqrt{2})$$$$$$$$z=e^{-\pi /4}{e}^{iln\sqrt{2}}=$$$$=e^{-\pi /4}[cos\left(ln\sqrt{2}\right)+isin\left(ln\sqrt{2}\right)]$$$$$$

Commented by mrW1 last updated on 05/Dec/17

$$verynice!$$

Commented by ajfour last updated on 05/Dec/17

$$indeed!$$

Commented by tawa tawa last updated on 05/Dec/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Answered by nnnavendu last updated on 05/Dec/17

$$Z={(1+i)}^i$$$$$$$$={(1+i)}^i\times 1^i\because 1^i=1$$$$={(1+i)}^{i^2}\because i^2=-1$$$$={(1+i)}^{-1}$$$$=\frac{1}{(1+i)}$$$$=\frac{1}{(1+i)}\times \frac{(1-i)}{(1-i)}$$$$=\frac{(1-i)}{(1^2-i^2)}\because i^2=-1$$$$=\frac{1-i}{1+1}$$$$=\frac{1-i}{2}$$$$=\frac{1}{2}-\frac{i}{2}$$$$realpart\frac{1}{2},imaginarypart\frac{-i}{2}$$

Commented by mrW1 last updated on 05/Dec/17

$$howdidyouget$$$$={(1+i)}^{i^2}$$$$from$$$$={(1+i)}^i$$$$?$$

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