Question

The real part of ${(1-\mathrm{i})}^{-\mathrm{i}}$ is

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

Answer 1

The correct option is A

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

Compute the real part:

Let $\mathrm{z}={(1-\mathrm{i})}^{-\mathrm{i}}$

Taking $\log$ on both sides,

$\log \left(\mathrm{z}\right)=-\mathrm{i}\log \left(1-\mathrm{i}\right)$ $\left[\log \left(a^x\right)=x\log \left(a\right)\right]$

$\Rightarrow$$\log \left(\mathrm{z}\right)=-\mathrm{i}\log \sqrt{2}\left(\cos \left(\frac{\mathrm{\pi }}{4}\right)-\mathrm{isin}\left(\frac{\mathrm{\pi }}{4}\right)\right)$

$\Rightarrow$$\log \left(\mathrm{z}\right)=-\mathrm{i}\log \sqrt{2}.{\mathrm{e}}^{-\mathrm{i}\frac{\mathrm{\pi }}{4}}$

$\Rightarrow$$\log \left(\mathrm{z}\right)=-\mathrm{i}\left(\frac{1}{2}\log \left(2\right)+\log \left({\mathrm{e}}^{-\mathrm{i}\frac{\mathrm{\pi }}{4}}\right)\right)$ $\left[\log \left(a.b\right)=\log \left(a\right)+\log \left(b\right)\right]$

$\Rightarrow$$\log \left(\mathrm{z}\right)=-\mathrm{i}\left[\frac{1}{2}\log \left(2\right)-\mathrm{i}\frac{\mathrm{\pi }}{4}\right]$

$\Rightarrow$$\log \left(\mathrm{z}\right)=\frac{-\mathrm{i}}{2}\log \left(2\right)-\frac{\mathrm{\pi }}{4}$

$\Rightarrow$ $\mathrm{z}={\mathrm{e}}^{-\frac{\mathrm{\pi }}{4}}{\mathrm{e}}^{-\frac{\mathrm{i}}{2}\log \left(2\right)}$

Taking real part ,

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

Hence, option (A) is the correct option.