Question Number 25152 by tawa tawa last updated on 05/Dec/17
$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{real}\:\mathrm{part}\:\mathrm{and}\:\mathrm{imaginary}\:\mathrm{part}\:\mathrm{of}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{number}:\:\:\:\:\mathrm{z}\:=\:\left(\mathrm{1}\:+\:\mathrm{i}\right)^{\mathrm{i}} \\ $$
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[ln((√2)e^(iπ/4) )] =i[ln(√2)+i(π/4)]=(−(π/4)+iln(√2)) z=e^(−π/4) e^(iln(√2)) = =e^(−π/4) [cos(ln(√2))+isin(ln(√2))]](https://www.tinkutara.com/question/Q25161.png)
$$\:\:\:\:{z}=\left(\mathrm{1}+{i}\right)^{{i}} \\ $$$${lnz}={iln}\left(\mathrm{1}+{i}\right)={i}\left[{ln}\left(\sqrt{\mathrm{2}}{e}^{{i}\pi/\mathrm{4}} \right)\right] \\ $$$$={i}\left[{ln}\sqrt{\mathrm{2}}+{i}\frac{\pi}{\mathrm{4}}\right]=\left(−\frac{\pi}{\mathrm{4}}+{iln}\sqrt{\mathrm{2}}\right) \\ $$$$ \\ $$$$\:\:{z}={e}^{−\pi/\mathrm{4}} {e}^{{iln}\sqrt{\mathrm{2}}} = \\ $$$$\:\:\:\:\:={e}^{−\pi/\mathrm{4}} \left[{cos}\left({ln}\sqrt{\mathrm{2}}\right)+{isin}\left({ln}\sqrt{\mathrm{2}}\right)\right] \\ $$$$ \\ $$
Commented by mrW1 last updated on 05/Dec/17
$${very}\:{nice}! \\ $$
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
$${Z}=\left(\mathrm{1}+{i}\right)^{{i}} \\ $$$$ \\ $$$$=\left(\mathrm{1}+{i}\right)^{{i}} ×\mathrm{1}^{{i}} \:\:\:\:\:\:\:\:\:\because\mathrm{1}^{{i}} =\mathrm{1} \\ $$$$=\left(\mathrm{1}+{i}\right)^{{i}^{\mathrm{2}} } \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\because{i}^{\mathrm{2}} =−\mathrm{1} \\ $$$$=\left(\mathrm{1}+{i}\right)^{−\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{1}+{i}\right)} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{1}+{i}\right)}×\frac{\left(\mathrm{1}−{i}\right)}{\left(\mathrm{1}−{i}\right)}\:\:\:\:\:\:\: \\ $$$$=\frac{\left(\mathrm{1}−{i}\right)}{\left(\mathrm{1}^{\mathrm{2}} −{i}^{\mathrm{2}} \right)}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\because{i}^{\mathrm{2}} =−\mathrm{1} \\ $$$$=\frac{\mathrm{1}−{i}}{\mathrm{1}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}−{i}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}−\frac{{i}}{\mathrm{2}} \\ $$$${real}\:{part}\frac{\mathrm{1}}{\mathrm{2}},{imaginary}\:{part}\frac{−{i}}{\mathrm{2}} \\ $$
Commented by mrW1 last updated on 05/Dec/17
$${how}\:{did}\:{you}\:{get} \\ $$$$=\left(\mathrm{1}+{i}\right)^{{i}^{\mathrm{2}} } \:\:\:\: \\ $$$${from} \\ $$$$=\left(\mathrm{1}+{i}\right)^{{i}} \\ $$$$? \\ $$