Question

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

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

Answer 1

The correct option is A $0$

$(1+i\sqrt{3}{)}^{12}=2^{12}{(\frac{1}{2}+i\frac{\sqrt{3}}{2})}^{12}$
$=2^{12}{\left(e^{i\frac{\pi }{3}}\right)}^{12}=2^{12}{e}^{4\pi i}$
$=2^{12}(\cos 4\pi +i\sin 4\pi )$
$=2^{12}(1+i\cdot 0)=a+i\cdot b$

Hence, $b=0$