Question

Find the real part of $(1-i{)}^i$.

• A. $Re\left(z\right)=e^{\frac{\pi }{4}}\cos \left(\frac{1}{2}\log 2\right)$
• B. $Re\left(z\right)=e^{-\frac{\pi }{4}}\cos \left(\frac{1}{2}\log 2\right)$
• C. $Re\left(z\right)=e^{-\frac{\pi }{4}}\cos \left(\log 2\right)$
• D. $Re\left(z\right)=e^{-\frac{\pi }{2}}\cos \left(\frac{1}{2}\log 2\right)$

Answer 1

Answer: (B)

$Re\left(z\right)=e^{-\frac{\pi }{4}}\cos \left(\frac{1}{2}\log 2\right)$

Given:

$z=(1-i{)}^i$
Taking $\log$ on both sides, we have
$\log z=i{\log }_e(1-i)$

$=i{\log }_e\sqrt{2}(\cos \frac{\pi }{4}-i\sin \frac{\pi }{4})$

$=i{\log }_e\left(\sqrt{2}{e}^{-i\frac{\pi }{4}}\right)$

$=i[\frac{1}{2}{\log }_{e}2+{\log }_e{e}^{-i\frac{\pi }{4}}]$

$=i[\frac{1}{2}{\log }_{e}2-i\frac{\pi }{4}]$

$=\frac{i}{2}{\log }_{e}2+\frac{\pi }{4}$

$\Rightarrow z=e^{\frac{\pi }{4}}{e}^{i\frac{\log 2}{2}}$

$\Rightarrow Re\left(z\right)=e^{\frac{\pi }{4}}\cos \left(\frac{1}{2}\log 2\right)$
Ans: A.