Question

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

• A. $r^{-\pi /4}\cos \left(\frac{1}{2}log2\right)$
• B. $-e^{-\pi /4}\sin \left(\frac{1}{2}log2\right)$
• C. $e^{\pi /4}\cos \left(\frac{1}{2}log2\right)$
• D. $e^{-\pi /4}\sin \left(\frac{1}{2}log2\right)$

Answer 1

The correct option is A $r^{-\pi /4}\cos \left(\frac{1}{2}log2\right)$

Let $z=(1-i{)}^{-i}$
On taking log on both sides, we get
$\Rightarrow logz=-ilog(1-i)$
$=-ilog\sqrt{2}\left(\cos \frac{\pi }{4}-i\sin \frac{\pi }{4}\right)$
$=-ilog(\sqrt{2}\cdot e^{-i\pi /4})$
$=-i\left[\frac{1}{2}log2+log{e}^{-i\pi /4}\right]$
$=-i\left[\frac{1}{2}log2-\frac{i\pi }{4}\right]=-\frac{i}{2}log2-\frac{\pi }{4}$
$\Rightarrow z=e^{-\pi /4}\cdot e^{-i/2log2}$
On taking real part only,
$\Rightarrow Re\left(z\right)=e^{-\pi /4}\cdot \cos \left(\frac{1}{2}log2\right)$.