In short

The square root of a complex number a + bi is another complex number x + yi such that (x + yi)^2 = a + bi. To find it, expand the left side, match real and imaginary parts, and solve the resulting system of two equations in x and y. Every non-zero complex number has exactly two square roots, and they are negatives of each other — just like \sqrt{4} = \pm 2 in the reals.

You know how to square a complex number. Take z = 3 + i and compute z^2 = (3 + i)^2 = 9 + 6i + i^2 = 8 + 6i. Squaring is straightforward — expand the binomial and use i^2 = -1.

The reverse question is harder. Given 8 + 6i, can you find a complex number whose square is 8 + 6i? That is, can you compute \sqrt{8 + 6i}?

You cannot just take the square root of each part — \sqrt{8} + \sqrt{6}\,i does not work. (Square it and check: (\sqrt{8} + \sqrt{6}\,i)^2 = 8 + 2\sqrt{48}\,i + 6i^2 = 2 + 2\sqrt{48}\,i, which is not 8 + 6i.) The real and imaginary parts of a square root are tangled together by the i^2 = -1 rule. Untangling them requires a method.

The method of comparison

The idea is clean. Suppose \sqrt{a + bi} = x + yi for some real numbers x and y you want to find. Square both sides:

(x + yi)^2 = a + bi

Expand the left side:

(x^2 - y^2) + 2xyi = a + bi

Now match real and imaginary parts. Two complex numbers are equal exactly when their real parts match and their imaginary parts match. This gives you two equations:

x^2 - y^2 = a \qquad \text{...(1)}
2xy = b \qquad \text{...(2)}

Two equations in two unknowns — a system you can solve.

There is a third equation hiding in plain sight. The modulus of both sides must match: |x + yi|^2 = |a + bi|, so x^2 + y^2 = \sqrt{a^2 + b^2}. Call this:

x^2 + y^2 = \sqrt{a^2 + b^2} \qquad \text{...(3)}

Now add equations (1) and (3):

2x^2 = a + \sqrt{a^2 + b^2} \implies x^2 = \frac{a + \sqrt{a^2 + b^2}}{2}

Subtract equation (1) from (3):

2y^2 = \sqrt{a^2 + b^2} - a \implies y^2 = \frac{\sqrt{a^2 + b^2} - a}{2}

Since \sqrt{a^2 + b^2} \geq |a|, both right-hand sides are non-negative, so real values of x and y always exist.

The signs of x and y are determined by equation (2): 2xy = b, which means x and y must have the same sign when b > 0 and opposite signs when b < 0. Pick one consistent choice; the other gives -x - yi, the second square root.

The method of comparison for finding square roots of complex numbersA flow diagram with three stages. Stage one: assume the square root is x plus yi. Stage two: square both sides and match real and imaginary parts to get two equations. Stage three: use the modulus equation to solve for x squared and y squared, then determine signs from the imaginary equation.Assume √(a+bi)= x + yiSquare both sides:x² − y² = a ...(1)2xy = b ...(2)x²+y² = √(a²+b²)...(3)(1)+(3) → x²(3)−(1) → y²(2) → signs of x,yThe modulus equation (3) is the key that unlocks the system
The three-step method of comparison. Set up the square root as $x + yi$, square to get two equations from matching parts, then use the modulus equation to decouple $x^2$ and $y^2$. The sign constraint from equation (2) completes the solution.

Square root of a complex number

If a + bi is a complex number with b \neq 0, then

\sqrt{a + bi} = \pm\left(\sqrt{\frac{\sqrt{a^2+b^2}+a}{2}} + \frac{b}{|b|}\sqrt{\frac{\sqrt{a^2+b^2}-a}{2}}\;i\right)

where \frac{b}{|b|} is +1 when b > 0 and -1 when b < 0. The two square roots are negatives of each other.

The formula looks heavy, and there is no reason to memorise it. The method of comparison — set \sqrt{a+bi} = x + yi, square, match, solve — is more reliable and just as fast. Use the formula only as a sanity check.

Finding \sqrt{a + bi} step by step

Here is the method applied to a specific number: find \sqrt{5 + 12i}.

Set up: Let \sqrt{5 + 12i} = x + yi.

Square: (x + yi)^2 = x^2 - y^2 + 2xyi = 5 + 12i.

Match parts:

Modulus: x^2 + y^2 = \sqrt{25 + 144} = \sqrt{169} = 13.

Solve for x^2 and y^2:

Determine signs: xy = 6 > 0, so x and y have the same sign. Either x = 3, y = 2 or x = -3, y = -2.

Result: \sqrt{5 + 12i} = \pm(3 + 2i).

Check: (3 + 2i)^2 = 9 + 12i + 4i^2 = 9 + 12i - 4 = 5 + 12i. Confirmed.

The two square roots of 5 plus 12i shown on the complex planeA complex plane with three key points. The number 5 plus 12i is marked at coordinates (5, 12) in the first quadrant. Its two square roots are at (3, 2) and (negative 3, negative 2), on opposite sides of the origin. Dashed lines connect the origin to each square root, showing they have equal distance from the origin but point in opposite directions.ReIm012345−1−2−312−1−23 + 2i−3 − 2i5 + 12i(off scale — shown for reference)the two square roots are diametrically opposite
The two square roots of $5 + 12i$ are $3 + 2i$ and $-3 - 2i$. They sit at diametrically opposite points — same distance from the origin, but pointing in opposite directions. Squaring either one produces $5 + 12i$.

Square root of a purely imaginary number

Take \sqrt{2i}. Here a = 0 and b = 2.

Let \sqrt{2i} = x + yi. Then x^2 - y^2 = 0 and 2xy = 2, so xy = 1.

From x^2 - y^2 = 0: x^2 = y^2, so |x| = |y|.

From xy = 1 > 0: x and y have the same sign.

So x = y. Substitute into xy = 1: x^2 = 1, giving x = 1. Then y = 1.

Result: \sqrt{2i} = \pm(1 + i).

Check: (1 + i)^2 = 1 + 2i + i^2 = 2i. Confirmed.

This reveals something about 1 + i: it is the square root of 2i. Its modulus is \sqrt{2} (since |1+i| = \sqrt{1+1} = \sqrt{2}), and its argument is 45°. Squaring doubles the argument to 90° and squares the modulus to 2, landing at 2i — which has modulus 2 and argument 90°. The geometry of squaring is: double the angle, square the distance.

Square root of a negative real number

For completeness: \sqrt{-9}. Here a = -9, b = 0.

The modulus is \sqrt{81 + 0} = 9.

x^2 = \frac{-9 + 9}{2} = 0, so x = 0.

y^2 = \frac{9 - (-9)}{2} = 9, so y = \pm 3.

Result: \sqrt{-9} = \pm 3i.

This matches the familiar result from Complex Numbers — Introduction: \sqrt{-9} = 3i.

Square roots of negative 9 on the imaginary axisA number line representing the complex plane. The real axis runs horizontally. The imaginary axis runs vertically. The number negative 9 is marked on the negative real axis. Its two square roots, 3i and negative 3i, are on the imaginary axis, equidistant from the origin.ReIm0−93i−3isquaring either rootgives −9 on the real axis
The two square roots of $-9$ are $3i$ and $-3i$, sitting on the imaginary axis. Squaring either one gives $(3i)^2 = 9i^2 = -9$. The square root of a negative real number always lies on the imaginary axis.

Applications

Solving quadratic equations with complex coefficients

The square root of a complex number appears directly when you use the quadratic formula on equations with complex coefficients. Take z^2 + 2iz - (3 + 4i) = 0.

The discriminant is D = (2i)^2 - 4(1)(-(3+4i)) = -4 + 12 + 16i = 8 + 16i.

You need \sqrt{8 + 16i}. Using the method of comparison:

Let \sqrt{8 + 16i} = x + yi. Then x^2 - y^2 = 8, 2xy = 16, so xy = 8.

Modulus: x^2 + y^2 = \sqrt{64 + 256} = \sqrt{320} = 8\sqrt{5}.

Adding: 2x^2 = 8 + 8\sqrt{5}, so x^2 = 4 + 4\sqrt{5}.

This is getting messy — the method works but the numbers are not clean. In practice, textbook problems are designed so the discriminant has a nice square root. The method itself is always the same; only the arithmetic varies.

Simplifying expressions involving \sqrt{-a} where a > 0

When you write \sqrt{-7}, you are finding the square root of a complex number (namely -7 + 0i). The method gives x = 0, y = \sqrt{7}, so \sqrt{-7} = i\sqrt{7}. This is the shortcut you learned earlier: for purely negative real numbers, the square root is i times the square root of the absolute value.

An interactive square-root explorer

Drag the point z around the complex plane. The readout shows z and the real and imaginary parts of z^2. Position z at various points and watch how squaring transforms the location — the angle doubles and the distance squares. To find a square root of a target, look for the point z whose z^2 readout matches your target.

Interactive square root explorerA coordinate plane representing the complex plane with the real axis horizontal and imaginary axis vertical, both ranging from negative 5 to 5. A red draggable point represents z. Readouts show z and z squared. The reader can position z to explore how squaring transforms complex numbers.ReIm1234−1−2−3−4↔ drag the red point
Drag $z$ around the plane. The top row shows $z$'s coordinates; the bottom row shows $z^2$. Place $z$ at $(3, 2)$ to see $z^2 = 5 + 12i$. Place $z$ at $(1, 1)$ to see $z^2 = 2i$. The square root of any target number is the $z$ whose readout matches it.

Worked examples

Example 1: Find $\sqrt{-7 + 24i}$

The numbers -7 and 24 are chosen so the algebra stays clean.

Step 1. Let \sqrt{-7 + 24i} = x + yi.

Why: you are looking for two real numbers x and y that assemble into the square root.

Step 2. Square both sides and match parts.

(x + yi)^2 = x^2 - y^2 + 2xyi = -7 + 24i

So x^2 - y^2 = -7 and 2xy = 24, giving xy = 12.

Why: the real parts give one equation and the imaginary parts give another. The factor of 2 in 2xy comes from the binomial expansion.

Step 3. Compute the modulus.

x^2 + y^2 = \sqrt{(-7)^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25

Why: |x + yi|^2 = x^2 + y^2 and |(-7 + 24i)|= 25. Since |(x+yi)^2| = |x+yi|^2, you get x^2 + y^2 = 25.

Step 4. Solve for x^2 and y^2.

Adding: 2x^2 = -7 + 25 = 18, so x^2 = 9 and x = \pm 3.

Subtracting: 2y^2 = 25 - (-7) = 32, so y^2 = 16 and y = \pm 4.

Why: adding equations (1) and (3) isolates x^2; subtracting isolates y^2. Both come out to perfect squares — a sign that the problem was designed with clean numbers.

Step 5. Determine signs. Since xy = 12 > 0, x and y share the same sign: (x, y) = (3, 4) or (-3, -4).

Result: \sqrt{-7 + 24i} = \pm(3 + 4i).

Square roots of negative 7 plus 24i on the complex planeA complex plane with three points. The number negative 7 plus 24i is in the second quadrant. Its two square roots 3 plus 4i and negative 3 minus 4i are on opposite sides of the origin, connected by a dashed line through the origin. The square root 3 plus 4i is in the first quadrant and negative 3 minus 4i is in the third quadrant.ReIm0123−1−2−312−1−23 + 4i−3 − 4iboth have modulus 55² = 25 = |−7 + 24i|
The two square roots of $-7 + 24i$ are $3 + 4i$ (first quadrant) and $-3 - 4i$ (third quadrant). Both have modulus $5$, and $5^2 = 25 = |-7 + 24i|$. Squaring either root doubles the argument angle and squares the distance, landing back at $-7 + 24i$.

Check: (3 + 4i)^2 = 9 + 24i + 16i^2 = 9 + 24i - 16 = -7 + 24i. The square root and the original number match.

Example 2: Find $\sqrt{-3 - 4i}$

The imaginary part is negative, so the sign constraint will force x and y to have opposite signs.

Step 1. Let \sqrt{-3 - 4i} = x + yi.

Why: same setup as before. The negative imaginary part will affect only the final sign determination.

Step 2. Square and match parts.

x^2 - y^2 = -3, \qquad 2xy = -4, \qquad xy = -2

Why: matching real parts gives the first equation, imaginary parts give 2xy = -4.

Step 3. Compute the modulus.

x^2 + y^2 = \sqrt{9 + 16} = \sqrt{25} = 5

Why: |-3 - 4i| = \sqrt{9 + 16} = 5.

Step 4. Solve.

Adding: 2x^2 = -3 + 5 = 2, so x^2 = 1 and x = \pm 1.

Subtracting: 2y^2 = 5 + 3 = 8, so y^2 = 4 and y = \pm 2.

Step 5. Since xy = -2 < 0, x and y have opposite signs. Either (x, y) = (1, -2) or (-1, 2).

Result: \sqrt{-3 - 4i} = \pm(1 - 2i).

Square roots of negative 3 minus 4i on the complex planeA complex plane with three points. The original number negative 3 minus 4i is in the third quadrant. Its two square roots 1 minus 2i and negative 1 plus 2i are on opposite sides of the origin. The root 1 minus 2i is in the fourth quadrant and negative 1 plus 2i is in the second quadrant.ReIm0123−1−212−1−2−3−4i (below frame)1 − 2i−1 + 2ixy = −2 < 0: opposite signsboth have modulus √5
The two square roots of $-3 - 4i$ are $1 - 2i$ (fourth quadrant) and $-1 + 2i$ (second quadrant). The negative imaginary part of $-3 - 4i$ forces the square roots to have $x$ and $y$ with opposite signs — one root in the second quadrant, the other in the fourth.

Check: (1 - 2i)^2 = 1 - 4i + 4i^2 = 1 - 4i - 4 = -3 - 4i. The sign of the imaginary part in the square root matters — getting it wrong would give -3 + 4i instead of -3 - 4i.

Common confusions

Going deeper

If you can find the square root of any complex number using the method of comparison, and you understand why there are always exactly two roots (negatives of each other), you have the essential content. The material below connects square roots to the polar form and to higher roots.

The polar shortcut

If you know the modulus r = |z| and the argument \theta = \arg(z), the square root has a clean polar form. Write z = r(\cos\theta + i\sin\theta). Then

\sqrt{z} = \sqrt{r}\left(\cos\frac{\theta}{2} + i\sin\frac{\theta}{2}\right)

and the second root is -\sqrt{z}, which corresponds to adding 180° to the half-angle.

For z = -7 + 24i: the modulus is 25, so \sqrt{r} = 5. The argument \theta satisfies \cos\theta = -7/25 and \sin\theta = 24/25. The half-angle gives \cos(\theta/2) and \sin(\theta/2), which work out to 3/5 and 4/5 respectively. So \sqrt{z} = 5(3/5 + 4/5\,i) = 3 + 4i. The same answer, reached through trigonometry instead of algebra.

This polar approach generalises to cube roots, fourth roots, and n-th roots of complex numbers — each obtained by dividing the argument by n and taking the n-th root of the modulus. For n-th roots, there are n roots equally spaced around a circle of radius r^{1/n}, separated by angles of 360°/n.

Connection to the quadratic formula

Every quadratic az^2 + bz + c = 0 has roots z = \frac{-b \pm \sqrt{D}}{2a}, where D = b^2 - 4ac is the discriminant. When a, b, c are real and D < 0, the discriminant is a negative real number, and \sqrt{D} is purely imaginary. But when the coefficients themselves are complex, D can be any complex number, and you need the full method of comparison to find \sqrt{D}.

This is the bridge between the quadratic formula and the complex square root: the formula works for all quadratics, but it requires you to take square roots of whatever the discriminant turns out to be — real, negative real, or genuinely complex. The method of comparison handles all three cases uniformly.

Uniqueness of the principal root

For positive real numbers, \sqrt{a} has a natural "principal" value: the positive root. For complex numbers, there is no universal agreement on which of the two roots is "the" square root. Some texts define the principal square root as the one with positive real part (or, if the real part is zero, positive imaginary part). This convention is consistent but not universally used. In most exam and textbook contexts, both roots are given with a \pm sign.

Where this leads next

The square root is a stepping stone to the geometric picture of complex numbers and to higher-degree equations.