Question

If ${(1+i\sqrt{3})}^{12}=a+ib$, here $a$ and $b$ are real, then the value of $b$ is

• A. ${\left(\sqrt{3}\right)}^{12}$
• B. ${\left(2\right)}^{12}$
• C. $0$
• D. $1$

Answer 1

Answer: (C)

$0$

We have, ${(1+i\sqrt{3})}^{12}=a+ib$
Now, ${\left[2(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3})\right]}^{12}=a+ib$ [De-Moivre's theorem]
$\Rightarrow 2^{12}{[\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}]}^{12}=a+ib$
$\Rightarrow 4096[\cos 4\pi +i\sin 4\pi ]=a+ib$
$\Rightarrow 4096[1+0]=a+ib$
$\Rightarrow 4096=a+ib$
On comparing, we get, $b=0$.