Q.Find the square root of the following complex number : $-8-6i$

To find the square root of a complex number $x+iy$, assume the (unknown) square root has the form $a+ib$ for real $a,b$, so $\sqrt{x+iy}=a+ib$. Squaring both sides gives $x+iy=(a+ib)^2=(a^2-b^2)+i(2ab)$, and equating real and imaginary parts on the two sides produces a pair of simultaneous real equations, $x=a^2-b^2$ and $y=2ab$, in the two unknowns $a$ and $b$. These are solved together — typically by using the auxiliary identity $(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2=x^2+y^2$, which gives $a^2+b^2$ directly as $\sqrt{x^2+y^2}$; combining this with $a^2-b^2=x$ then pins down $a^2$ and $b^2$ separately by simple addition and subtraction, and taking square roots gives $a$ and $b$ up to sign. The signs of $a$ and $b$ are not independent: because $y=2ab$ fixes the relative sign of $a$ and $b$ (same sign if $y>0$, opposite if $y<0$), only two of the four sign combinations from $a=\pm\sqrt{a^2}$ and $b=\pm\sqrt{b^2}$ actually satisfy the original equation — so a complex number always has exactly two square roots, and they are negatives of each other, matching the familiar fact that every nonzero number (real or complex) has two square roots.