Question

Find the square root of complex number $-8-6i$

Answer 1

Let the square root be $x+iy$

$\Rightarrow (x+iy)=\sqrt{-8-6i}$

Taking square on both the side

$\Rightarrow x^2-y^2+2xyi=-8-6i$

By comparing we get.

$\Rightarrow x^2-y^2=8$ and $2xy=-6$

$\Rightarrow xy=-3$

$\Rightarrow y=\frac{-3}{x}$

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

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

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

$\Rightarrow x^4+8x^2-9=0$

$\Rightarrow x^4+9x^2-x^2-9=0$

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

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

$\Rightarrow x^2=-9;x^2=1$

Since $x$ is real we get $x=\pm 1$

Taking $x=\pm 1$ we get $y=-3$

$\Rightarrow$ square root is $1-3i$

Taking $x=-1$ we get $y=3$

$\Rightarrow$ square root is $-1+3i$

Square root is $\pm (1-3i)$