$\sqrt{-8-6\mathrm{i}} \mathrm{is} \mathrm{equal} \mathrm{to}$

The given complex number is $\sqrt{-8-6i}$

Suppose that $\sqrt{-8-6i}=x+iy$

Squaring on both sides 

$-8-6i=x^2-y^2+2ixy$

Comparing the real and imaginary parts on both sides 

$x^2-y^2=-8, xy=-3$

The possible values of $x,y$ are $x=1,y=-3$

Hence, $\sqrt{-8-6i}=\boxed{\pm \left(1-3i\right)}$

$$\begin{gathered} \boxed{\text{ SQUARE ROOT OF COMPLEX NUMBER } \\ \sqrt{a+ib}=\pm \left\{\frac{\sqrt{\left|z\right|+a}}{2}+i\frac{\sqrt{\left|z\right|-a}}{2}\right\}\text{ for }b>0 \\ =\pm \left\{\frac{\sqrt{\left|z\right|+a}}{2}-i\frac{\sqrt{\left|z\right|-a}}{2}\right\}\text{ for }b<0\text{ where }\left|z\right|=\sqrt{a^2+b^2}} \end{gathered}$$