Square Roots of a Complex Number
Every complex number has a square root in $\mathbb{C}$. Given $a + bi$, we find $x + yi$ such that $(x + yi)^2 = a + bi$ by solving a system of two real equations. The result is a pair of opposite complex numbers $\pm(x + yi)$.
Quick Reference
| Quantity | Formula (when $b \neq 0$) |
|---|---|
| $\sqrt{a + bi}$ | $\pm!\left(\sqrt{\dfrac{a + \sqrt{a^2+b^2}}{2}} + i\cdot\dfrac{b}{ |
| $x^2$ | $\dfrac{a + \sqrt{a^2+b^2}}{2}$ |
| $y^2$ | $\dfrac{-a + \sqrt{a^2+b^2}}{2}$ |
| Sign condition | $xy$ and $b$ have the same sign |
Deriving the Square Root Formula
Given a complex number $a + bi$, we seek $x + yi$ such that:
$ (x + yi)^2 = a + bi $Expanding the left side:
$ (x + yi)^2 = x^2 - y^2 + 2xyi $Matching real and imaginary parts gives a system of two equations:
$ \left\{ \begin{aligned} x^2 - y^2 &= a \\ 2xy &= b \end{aligned} \right. $To solve this, we use the identity $(x^2 + y^2)^2 = (x^2 - y^2)^2 + 4x^2y^2$:
$ (x^2 + y^2)^2 = a^2 + b^2 $Since $x^2 + y^2 \geq 0$, we take the non-negative square root:
$ x^2 + y^2 = \sqrt{a^2 + b^2} $Combining this with $x^2 - y^2 = a$:
$ x^2 = \frac{a + \sqrt{a^2 + b^2}}{2}, \qquad y^2 = \frac{-a + \sqrt{a^2 + b^2}}{2} $Both $x^2$ and $y^2$ are non-negative regardless of the sign of $a$, so real solutions $x$ and $y$ always exist.
The solutions come in pairs $\pm(x, y)$. Since $2xy = b$, the product $xy$ must have the same sign as $b$. This determines which combination of signs to choose.
Square Root Formula. For $a + bi$ with $b \neq 0$:
$ \sqrt{a + bi} = \pm\left(\sqrt{\frac{a + \sqrt{a^2+b^2}}{2}} + i\cdot\frac{b}{|b|}\sqrt{\frac{-a + \sqrt{a^2+b^2}}{2}}\right) $For the case $b = 0$:
$ \sqrt{a} = \begin{cases} \pm\sqrt{a} & \text{if } a \geq 0 \\ \pm\,i\sqrt{-a} & \text{if } a < 0 \end{cases} $
Properties of Complex Square Roots
Every non-zero complex number has exactly two square roots, which are negatives of each other (i.e., $\pm r$ for some $r \in \mathbb{C}$). These two values coincide only if $a + bi = 0$.
- If $b = 0$ and $a > 0$: the square roots are real, $\pm\sqrt{a}$.
- If $b = 0$ and $a < 0$: the square roots are purely imaginary, $\pm\,i\sqrt{-a}$.
- If $b \neq 0$: the square roots are genuinely complex (neither real nor purely imaginary).
Example 1. Find $\sqrt{3 + 4i}$.
Solution. Here $a = 3$ and $b = 4$.
Compute: $\sqrt{a^2 + b^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
Then:
$ x^2 = \frac{3 + 5}{2} = 4, \quad x = 2 $$ y^2 = \frac{-3 + 5}{2} = 1, \quad y = 1 $Since $b = 4 > 0$, we need $xy > 0$, so $x$ and $y$ have the same sign. The two square roots are:
$ \sqrt{3 + 4i} = \pm(2 + i) $Verification: $(2 + i)^2 = 4 + 4i + i^2 = 4 + 4i - 1 = 3 + 4i$. Correct.
Example 2. Find $\sqrt{-5 + 12i}$.
Solution. Here $a = -5$ and $b = 12$.
Compute: $\sqrt{a^2 + b^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.
Then:
$ x^2 = \frac{-5 + 13}{2} = 4, \quad x = 2 $$ y^2 = \frac{5 + 13}{2} = 9, \quad y = 3 $Since $b = 12 > 0$, we need $xy > 0$, so $x$ and $y$ have the same sign:
$ \sqrt{-5 + 12i} = \pm(2 + 3i) $Verification: $(2 + 3i)^2 = 4 + 12i + 9i^2 = 4 + 12i - 9 = -5 + 12i$. Correct.
Quadratic Equations with Complex Coefficients
The quadratic formula
$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $still works when the coefficients $a$, $b$, $c$ are complex numbers. When $b^2 - 4ac$ is complex (not real), the square root in the formula must be understood as one of the two complex square roots computed above.
Example 3. Solve $x^2 - (3 + 2i)x + (5 + i) = 0$.
Solution. Here $a = 1$, $b = -(3 + 2i)$, $c = 5 + i$.
Compute the discriminant:
$ \begin{aligned} b^2 - 4ac &= (-(3+2i))^2 - 4(5+i) \\ &= (9 + 12i + 4i^2) - (20 + 4i) \\ &= (9 + 12i - 4) - 20 - 4i \\ &= 5 + 12i - 20 - 4i \\ &= -15 + 8i \end{aligned} $Find $\sqrt{-15 + 8i}$: here $a = -15$, $b = 8$, $\sqrt{a^2+b^2} = \sqrt{225 + 64} = \sqrt{289} = 17$.
$ x^2 = \frac{-15 + 17}{2} = 1, \quad x = 1 $$ y^2 = \frac{15 + 17}{2} = 16, \quad y = 4 $Since $b = 8 > 0$, take $xy > 0$: $\sqrt{-15 + 8i} = \pm(1 + 4i)$.
The two solutions are:
$ \begin{aligned} x_1 &= \frac{(3 + 2i) + (1 + 4i)}{2} = \frac{4 + 6i}{2} = 2 + 3i \\ x_2 &= \frac{(3 + 2i) - (1 + 4i)}{2} = \frac{2 - 2i}{2} = 1 - i \end{aligned} $Frequently Asked Questions
How many square roots does a complex number have?
Every non-zero complex number has exactly two square roots, and they are negatives of each other. For example, the square roots of $3 + 4i$ are $2 + i$ and $-(2 + i) = -2 - i$. The number zero has only one square root: zero.
