Square Roots of a Complex Number

To find the square roots of $a+ib$, set $x+iy$ to be a square root, so $a+ib=(x+iy)^2$ for real $x,y$. Expanding,

Equating real and imaginary parts: $x^2-y^2=a$ and $2xy=b$. To pin down $x^2+y^2$ as well, note $(x^2+y^2)^2=(x^2-y^2)^2+(2xy)^2=a^2+b^2$, so, since $x^2+y^2$ is positive,

Now solving the pair $x^2-y^2=a$ and $x^2+y^2=|z|$ simultaneously:

Since $2xy=b$: $x$ and $y$ must have the same sign when $b>0$, and opposite signs when $b<0$ (this is what fixes which combination of the two $\pm$ choices is valid — they are not independent). Writing $\operatorname{sgn}(b)=b/|b|$ for $b\ne0$, the formula for the square root of a complex number is

and the overall $\pm$ reflects that both $w$ and $-w$ square to the same $a+ib$ whenever $w^2=a+ib$. …