Question

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

Answer 1

Given that $(\sqrt{8}+i{)}^{50}=3^{49}(a+ib)$
Taking modulus and squaring both sides, we get
$(8+1{)}^{50}=3^{98}(a^2+b^2)$
$\Rightarrow 9^{50}=3^{98}(a^2+b^2)$
$\Rightarrow 3^{100}=3^{98}(a^2+b^2)$
$\Rightarrow a^2+b^2=9$