Question

Simplify:$\sqrt{8+6i}$

Answer 1

(A) $8+6i=(a+bi{)}^2$

$=(a^2-b^2)+2abi$

So $a^2-b^2=8$ …………$\left(1\right)$

$2ab=6$ …………..$\left(2\right)$

So $b=\frac{3}{a}$

In $\left(1\right)$

$a^2-{\left(\frac{3}{a}\right)}^2=8$

$\Rightarrow a^2-\frac{9}{a^2}=8$

$\Rightarrow a^4-9=8a^2$

$\Rightarrow a^4-8a^2-9=0$

$\Rightarrow (a^2-9)(a^2+1)=0$

$a^2$ can't be negative

$a^2=9$

$a=\pm 3$

$b=\pm 1$

Square root is $(\pm 3\pm i)$.