Question and Answers Forum

All Questions      Topic List

Number Theory Questions

Previous in All Question      Next in All Question      

Previous in Number Theory      Next in Number Theory      

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

What is the real part and imaginary part of the complex number: z = (1 + i)^i

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

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

please help.

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

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))]

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

Commented by mrW1 last updated on 05/Dec/17

very nice!

$${very}\:{nice}! \\ $$

Commented by ajfour last updated on 05/Dec/17

indeed!

$${indeed}! \\ $$

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

God bless you sir.

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

Answered by nnnavendu last updated on 05/Dec/17

Z=(1+i)^i =(1+i)^i ×1^i ∵1^i =1 =(1+i)^i^2 ∵i^2 =−1 =(1+i)^(−1) =(1/((1+i))) =(1/((1+i)))×(((1−i))/((1−i))) =(((1−i))/((1^2 −i^2 ))) ∵i^2 =−1 =((1−i)/(1+1)) =((1−i)/2) =(1/2)−(i/2) real part(1/2),imaginary part((−i)/2)

$${Z}=\left(\mathrm{1}+{i}\right)^{{i}} \\ $$$$ \\ $$$$=\left(\mathrm{1}+{i}\right)^{{i}} ×\mathrm{1}^{{i}} \:\:\:\:\:\:\:\:\:\because\mathrm{1}^{{i}} =\mathrm{1} \\ $$$$=\left(\mathrm{1}+{i}\right)^{{i}^{\mathrm{2}} } \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\because{i}^{\mathrm{2}} =−\mathrm{1} \\ $$$$=\left(\mathrm{1}+{i}\right)^{−\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{1}+{i}\right)} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{1}+{i}\right)}×\frac{\left(\mathrm{1}−{i}\right)}{\left(\mathrm{1}−{i}\right)}\:\:\:\:\:\:\: \\ $$$$=\frac{\left(\mathrm{1}−{i}\right)}{\left(\mathrm{1}^{\mathrm{2}} −{i}^{\mathrm{2}} \right)}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\because{i}^{\mathrm{2}} =−\mathrm{1} \\ $$$$=\frac{\mathrm{1}−{i}}{\mathrm{1}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}−{i}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}−\frac{{i}}{\mathrm{2}} \\ $$$${real}\:{part}\frac{\mathrm{1}}{\mathrm{2}},{imaginary}\:{part}\frac{−{i}}{\mathrm{2}} \\ $$

Commented by mrW1 last updated on 05/Dec/17

how did you get =(1+i)^i^2 from =(1+i)^i ?

$${how}\:{did}\:{you}\:{get} \\ $$$$=\left(\mathrm{1}+{i}\right)^{{i}^{\mathrm{2}} } \:\:\:\: \\ $$$${from} \\ $$$$=\left(\mathrm{1}+{i}\right)^{{i}} \\ $$$$? \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com