A) \[a=2,b=-1\]
B) \[a=1,b=0\]
C) \[a=0,b=1\]
D) \[a=-1,b=2\]
Correct Answer: B
Solution :
Given, \[{{\left( \frac{1-i}{1+i} \right)}^{100}}=a+ib\]; \[\left[ \left( \frac{1-i}{1+i} \right)\times \left( \frac{1-i}{1-i} \right) \right]=a+ib\] Þ \[a+ib={{\left[ \frac{{{(1-i)}^{2}}}{2} \right]}^{100}}={{\left[ \frac{-2i}{2} \right]}^{100}}={{(-i)}^{100}}\] Þ \[a+ib={{\left[ {{(i)}^{4}} \right]}^{25}}=1+0i,\]Hence, \[a=1,b=0\].
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