• # question_answer If  ${{\left( \frac{1-i}{1+i} \right)}^{100}}=a+ib$,  then [MP PET 1998] A) $a=2,b=-1$ B) $a=1,b=0$ C) $a=0,b=1$ D) $a=-1,b=2$

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, $a=1,b=0$.