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