A) \[a=1,b=0\]
B) \[a=0,b=1\]
C) \[a=-1,b=2\]
D) \[a=2,b=-1\]
Correct Answer: A
Solution :
Given, \[{{\left( \frac{1-i}{1+i} \right)}^{100}}=a+ib\] \[\frac{1-i}{1+i}=\frac{{{(1-i)}^{2}}}{(1+i)(1-i)}=\frac{{{1}^{2}}+{{i}^{2}}-2i}{1-{{i}^{2}}}\] \[=\frac{1-1-2i}{1+1}=-i\] \[\therefore \] \[{{\left( \frac{1-i}{1+i} \right)}^{100}}={{(-i)}^{100}}\] \[={{[{{(-i)}^{2}}]}^{50}}\] \[={{[-i]}^{50}}\] \[[\because -{{(i)}^{2}}=-1]\] \[=1\] \[=1+0i\] \[\therefore \] \[a=1,b=0\]
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