MCQ
If ${(1 + i\sqrt 3 )^9} = a + ib,$ then $b$ is equal to
  • A
    $1$
  • B
    $256$
  • $0$
  • D
    ${9^3}$

Answer

Correct option: C.
$0$
c
(c) $1 + i\sqrt 3 = 2\left( {\frac{1}{2} + i\frac{{\sqrt 3 }}{2}} \right) = 2\left[ {\cos \frac{\pi }{3} + i\sin \frac{\pi }{3}} \right] = 2{e^{i\pi /3}}$
 ${(1 + i\sqrt 3 )^9} = {(2{e^{i\pi /3}})^9} = {2^9}.{e^{i(3\pi )}}$
$ = {2^9}(\cos 3\pi + i\sin 3\pi ) = - {2^9}$
$a + ib = {(1 + i\sqrt 3 )^9} = - {2^9}$; 

$b = 0$.

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