MCQ
If ${\left( {\frac{{1 - i}}{{1 + i}}} \right)^{100}} = a + ib$, then
  • A
    $a = 2,b = - 1$
  • $a = 1,b = 0$
  • C
    $a = 0,b = 1$
  • D
    $a = - 1,b = 2$

Answer

Correct option: B.
$a = 1,b = 0$
b
(b) Given, ${\left( {\frac{{1 - i}}{{1 + i}}} \right)^{100}} = a + ib$; $\left[ {\left( {\frac{{1 - i}}{{1 + i}}} \right) \times \left( {\frac{{1 - i}}{{1 - i}}} \right)} \right] = a + ib$
==> $a + ib = {\left[ {\frac{{{{(1 - i)}^2}}}{2}} \right]^{100}} = {\left[ {\frac{{ - 2i}}{2}} \right]^{100}} = {( - i)^{100}}$
==> $a + ib = {\left[ {{{(i)}^4}} \right]^{25}} = 1 + 0i,$Hence, $a = 1,b = 0$.

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