Question

If ${\left(\sqrt{8}+i\right)}^{50}=3^{49}\left(a+ib\right)$, then $a^2+b^2$ is

• A. $3$
• B. $8$
• C. $9$
• D. $\sqrt{8}$

Answer 1

Answer: (C)

Writing $\sqrt{8}+i$ in polar form,
$\sqrt{8}+i=r\left(cos\theta +isin\theta \right)$, where $r=\sqrt{{\left(\sqrt{8}\right)}^2+1^2}=3$
and $tan\theta =\frac{1/3}{\sqrt{8}/3}=\frac{1}{\sqrt{8}}$
$\therefore \theta =ta{n}^{-1}\frac{1}{2\sqrt{2}}$
Now, ${\left(\sqrt{8}+i\right)}^{50}=3^{50}{\left(cos\theta +isin\theta \right)}^{50}$
$=3^{50}\left(cos50\theta +isin50\theta \right)$
But given that ${\left(\sqrt{8}+i\right)}^{50}=3^{49}\left(a+ib\right)$
$\therefore a=3cos50\theta ,b=3sin50\theta$
$\therefore a^2+b^2=9\left(co{s}^{2}50\theta +si{n}^{2}50\theta \right)=9$