Complex Numbers And Quadratic Equations question 440
Question: If $ {{( \frac{1-i}{1+i} )}^{100}}=a+ib $ , then [MP PET 1998]
Options:
A) $ a=2,b=-1 $
B) $ a=1,b=0 $
C) $ a=0,b=1 $
D) $ a=-1,b=2 $
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Answer:
Correct Answer: B
Solution:
Given, $ {{( \frac{1-i}{1+i} )}^{100}}=a+ib $ ; $ [ ( \frac{1-i}{1+i} )\times ( \frac{1-i}{1-i} ) ]=a+ib $
Þ $ a+ib={{[ \frac{{{(1-i)}^{2}}}{2} ]}^{100}}={{[ \frac{-2i}{2} ]}^{100}}={{(-i)}^{100}} $
Þ $ a+ib={{[ {{(i)}^{4}} ]}^{25}}=1+0i, $ Hence, $ a=1,b=0 $ .
BETA
