Square roots of a complex number

"Square root of $w$" means a complex number whose square is $w$. Unlike $\sqrt2$, this is not a new symbol to invent: it is a small algebra problem.

Solving $z^2 = w$ is a quadratic equation in $z$. A quadratic has two solutions, counted with repeats.

If $z_0$ is one solution, then $-z_0$ is the other, because $(-z_0)^2 = z_0^2 = w$. The only exception is $w = 0$, where the two roots coincide.

So the two square roots are always opposite points on the Argand diagram: same distance from the origin, directions differing by a half turn. An answer with only one root is incomplete.

Write the unknown as $z = x + iy$ with $x$ and $y$ real, and square it:

$z^2 = (x + iy)^2 = x^2 + 2ixy + i^2y^2 = (x^2 - y^2) + 2ixy$

This has to equal $w = a + ib$. Two complex numbers are equal only when both parts match, so

$x^2 - y^2 = a \qquad\text{and}\qquad 2xy = b$

That is two equations for two unknowns, but they are not linear, and they are awkward on their own. The trick that makes them easy is the modulus.

Since $\lvert z^2 \rvert = \lvert z \rvert^2$, and $z^2 = w$,

$x^2 + y^2 = \lvert z \rvert^2 = \lvert w \rvert = \sqrt{a^2 + b^2}$

Now $x^2 - y^2$ and $x^2 + y^2$ are known, so adding gives $2x^2$ and subtracting gives $2y^2$. Both $x^2$ and $y^2$ follow, and then the sign of $xy$ picks the correct pair of signs.

1. Write $x^2 - y^2 = a$ and $2xy = b$.
2. Compute $x^2 + y^2 = \sqrt{a^2 + b^2}$.
3. Add and subtract the equations for $x^2 - y^2$ and $x^2 + y^2$ to get $x^2$ and $y^2$. Then take square roots to get the sizes of $x$ and $y$, which are $\lvert x \rvert$ and $\lvert y \rvert$.
4. Use the sign of $b$ to decide the signs:
   • if $b > 0$, then $xy > 0$, so $x$ and $y$ have the same sign;
   • if $b < 0$, then $xy < 0$, so $x$ and $y$ have opposite signs;
   • if $b = 0$, one of $x$, $y$ is zero.

Only two of the four sign combinations work. The other two give $xy$ the wrong sign, so they don't satisfy $2xy = b$.