Square Root of a Complex Number

Here |z| = √(a²+b²) and sgn(b) = +1 if b ≥ 0, −1 if b < 0. The two square roots are always negatives of each other. Derived by setting (x+iy)² = a+ib and solving the system x²−y²=a, 2xy=b using x²+y²=|z|.

Every non-zero complex number has exactly two square roots, and they are negatives of each other. Finding them requires only algebra and the definition of modulus.

Setup

Let $z = a + ib$. We want to find $w = x + iy$ such that $w^2 = z$.

Squaring $w$:

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

Equating real and imaginary parts with $a + ib$:

$$x^2 - y^2 = a \tag{1}$$ $$2xy = b \tag{2}$$

This is a system of two equations in two unknowns. But it is not linear — equation (1) involves a difference of squares. The key is to introduce a third equation using the modulus.

Introducing the Modulus

Compute $|w|^2 = x^2 + y^2$. Since $w^2 = z$, we have $|w^2| = |z|$, so $|w|^2 = |z|$:

$$x^2 + y^2 = \sqrt{a^2 + b^2} = |z| \tag{3}$$

Now equations (1) and (3) form a clean linear system in $x^2$ and $y^2$:

$$(1) + (3): \quad 2x^2 = |z| + a \implies x^2 = \frac{|z| + a}{2}$$ $$(3) - (1): \quad 2y^2 = |z| - a \implies y^2 = \frac{|z| - a}{2}$$

Determining the Signs

Taking square roots:

$$|x| = \sqrt{\frac{|z| + a}{2}}, \qquad |y| = \sqrt{\frac{|z| - a}{2}}$$

The signs of $x$ and $y$ are not independent — they are linked by equation (2): $2xy = b$.

Choose $x > 0$ (the case $x < 0$ gives the other root, which is just $-w$). Then from (2):

$$\operatorname{sgn}(y) = \operatorname{sgn}(b)$$

So:

$$y = \operatorname{sgn}(b)\sqrt{\frac{|z| - a}{2}}$$

The Result

$$\sqrt{a + ib} = \pm\left(\sqrt{\frac{|z| + a}{2}} + i\operatorname{sgn}(b)\sqrt{\frac{|z| - a}{2}}\right)$$

where $|z| = \sqrt{a^2 + b^2}$. The $\pm$ gives both roots.

Special Cases

When $b = 0$ (real number): The formula gives $\sqrt{a} + 0i$ for $a > 0$ and $0 + i\sqrt{|a|}$ for $a < 0$ — consistent with what we already know.

When $b > 0$: Both $x$ and $y$ are positive — the root lies in the first quadrant.

When $b < 0$: $x$ is positive but $y$ is negative — the root lies in the fourth quadrant.

The two square roots always lie in opposite quadrants (or on opposite half-axes), as they must since one is the negative of the other.