Question

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

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

Answer 1

Answer: (A)

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