Complex Numbers And Quadratic Equations question 841
Question: The real part of $ {{(1-i)}^{-i}} $ is [RPET 1999]
Options:
A) $ {e^{-\pi /4}}\cos ( \frac{1}{2}\log 2 ) $
B) $ -{e^{-\pi /4}}\sin ( \frac{1}{2}\log 2 ) $
C) $ {e^{\pi /4}}\cos ( \frac{1}{2}\log 2 ) $
D) $ {e^{-\pi /4}}\sin ( \frac{1}{2}\log 2 ) $
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Answer:
Correct Answer: A
Solution:
Let $ z={{(1-i)}^{-i}} $ . Taking log on both sides,
$ \Rightarrow \log z $ $ =-i\log (1-i) $ $ =-i\log \sqrt{2}( \cos \frac{\pi }{4}-i\sin \frac{\pi }{4} ) $ $ =-i\log ( \sqrt{2}{e^{-i\pi /4}} ) $ $ =-i[ \frac{1}{2}\log 2+\log {e^{-i\pi /4}} ] $ $ =-i[ \frac{1}{2}\log 2-\frac{i\pi }{4} ] $ $ =-\frac{i}{2}\log 2-\frac{\pi }{4} $
Þ $ z={e^{-\pi /4}}{e^{-i/2\log 2}} $ . Taking real part only,
$ \Rightarrow Re(z)={e^{-\pi /4}}\cos ( \frac{1}{2}\log 2 ) $ .
BETA
