Question

If $(\sqrt{3}+i{)}^{100}=2^{99}(a+ib)$, then $a^2+b^2$ is equal to

• A. $\sqrt{2}$
• B. $4$
• C. $\sqrt{3}$
• D. none of these

Answer 1

Answer: (B)

Since ${\left(\sqrt{3}+i\right)}^{100}=2^{99}\left(a+ib\right)$
$\therefore {\left(\sqrt{3}-i\right)}^{100}=2^{99}\left(a-ib\right)$
Multiplying, we get
${\left(\sqrt{3}+i\right)}^{100}{\left(\sqrt{3}-i\right)}^{100}=2^{198}\left(a^2+b^2\right)$
$\therefore a^2+b^2=\frac{{\left(3+1\right)}^{100}}{2^{198}}=\frac{2^{200}}{2^{198}}$
$=2^2=4$.