The real part of (1 - i)^ - i is - Vidyadip

MCQ

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

✓
${e^{ - \pi /4}}\cos \left( {\frac{1}{2}\log 2} \right)$
B
$ - {e^{ - \pi /4}}\sin \left( {\frac{1}{2}\log 2} \right)$
C
${e^{\pi /4}}\cos \left( {\frac{1}{2}\log 2} \right)$
D
${e^{ - \pi /4}}\sin \left( {\frac{1}{2}\log 2} \right)$

✓

Answer

Correct option: A.

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

a
(a)Let $z = {(1 - i)^{ - i}}$. Taking $ log $ on both sides,
$ \Rightarrow \,\log \,z$$ = - i\,\,\log (1 - i)$$ = - i\,\log \sqrt 2 \,\left( {\cos \frac{\pi }{4} - i\sin \frac{\pi }{4}} \right)$
$ = - \,i\,\log \left( {\sqrt 2 {e^{ - \,i\,\pi /4}}} \right)$$ = - i\,\left[ {\frac{1}{2}\log 2 + \log \,{e^{ - i\,\pi /4}}} \right]$
$ = - i\,\left[ {\frac{1}{2}\log 2 - \frac{{i\pi }}{4}} \right]$ $ = - \frac{i}{2}\log \,2\, - \frac{\pi }{4}$
==> $z = {e^{ - \pi /4}}\,\,{e^{ - i/2\,\log 2}}$. Taking real part only,
$ \Rightarrow \,\,{\mathop{\rm Re}\nolimits} (z) = \,{e^{ - \pi /4}}\,\cos \,\left( {\frac{1}{2}\log 2} \right)$.

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