Complex Numbers

Updated 5 October 2026

A complex number has a real part and an imaginary part, and is written a + bi. The imaginary unit i is defined by $i^2=-1$i² = −1. Such a number does not fit on a line, so complex numbers live on a plane, where adding is a shift and multiplying is a turn.

What is the square root of a negative number?

A negative number has no real square root, because every real number squared is zero or positive, so $x^2=-1$x² = −1 has no solution on the number line. Mathematics answers this by defining a new number, i, whose square is −1. Every negative number then has two square roots: those of −16 are 4i and −4i, because squaring 4i gives $16\times i^2$16 × i², and 16 × (−1) = −16. In general $\sqrt{-a}=i\sqrt{a}$√(−a) = i√a for any positive a, so $\sqrt{-25}=5i$√(−25) = 5i. The definition works because i obeys all the ordinary rules of algebra, with one extra fact: $i^2=-1$i² = −1 wherever it appears. One rule does change. The law $\sqrt{a}\times \sqrt{b}=\sqrt{ab}$√a × √b = √(ab) holds only when a and b are not negative, so $\sqrt{-1}\times \sqrt{-1}$√(−1) × √(−1) is i × i = −1, not $\sqrt{1}=1$√1 = 1. Check a square root by squaring it: squaring 5i gives $25\times i^2$25 × i², and 25 × (−1) = −25, the number you started from.

With i defined by $i^2=-1$i² = −1, every other square root of a negative follows: $\sqrt{-9}=\sqrt{9}\times \sqrt{-1}=3i$√(−9) = √9 × √(−1) = 3i. Algebra still works with i in an expression: expand brackets as before, and replace $i^2$i² with −1 wherever it appears. See The Imaginary Unit.

The powers of i repeat in a cycle of four: i, −1, −i, 1. To find i²⁰²⁷, divide the exponent by 4 and keep the remainder. Since 2027 = 4 × 506 + 3, i²⁰²⁷ $=i^3=-i$= i³ = −i. See Powers of i.

How do you add and multiply complex numbers?

To add complex numbers, add the real parts and the imaginary parts separately: (3 + 2i) + (1 − 5i) = 4 − 3i, since 3 + 1 = 4 and 2 − 5 = −3. See Adding Complex Numbers. For multiplication, expand the brackets as if i were a letter, then replace $i^2$i² with −1: (2 + i)(3 + 4i) gives $6+8i+3i+4i^2$6 + 8i + 3i + 4i², which is 2 + 11i. Adding never mixes the parts, because a real number and a multiple of i are unlike terms, as 3 and 2x are. Multiplying does mix them, because i × i is real. Division needs one more step: multiply the top and bottom by the conjugate of the denominator, which makes the denominator real. Moduli multiply, which checks the product: $|2+i{|}^2$|2 + i|² is 5, $|3+4i{|}^2$|3 + 4i|² is 25, and $|2+11i{|}^2$|2 + 11i|² is 4 + 121 = 125 = 5 × 25.

Written out in full, with $i^2$i² replaced by −1:

$\left(3+2i\right)\left(1-5i\right)=3-15i+2i-10i^2=3-13i+10=13-13i$(3 + 2i)(1 − 5i) = 3 − 15i + 2i − 10i² = 3 − 13i + 10 = 13 − 13i.

The last step is where mistakes happen: $-10i^2$−10i² is −10 × (−1) = +10, not −10. See Multiplying Complex Numbers.

Dividing with the conjugate

The conjugate of a + bi is a − bi. A number times its conjugate is always real, because the middle terms cancel and $-b^2{i}^2$−b²i² becomes $+b^2$+b²: $\left(a+\text{bi}\right)\left(a-\text{bi}\right)=a^2+b^2$(a + bi)(a − bi) = a² + b². See Complex Conjugates.

It is the same move as rationalizing a denominator with a surd in it. See Dividing Complex Numbers.

(4 + i) / (2 − 3i) = (4 + i)(2 + 3i) / ((2 − 3i)(2 + 3i)) = (5 + 14i) / 13.

Where do complex numbers come from in an equation?

Complex numbers appear when a quadratic has a negative discriminant. Its graph then misses the x-axis, so there are no real roots, yet the quadratic formula still works once $\sqrt{-1}$√(−1) is written as i. For $x^2-4x+13=0$x² − 4x + 13 = 0, the discriminant $b^2-4ac$b² − 4ac is 16 − 52 = −36, and $\sqrt{-36}=6i$√(−36) = 6i, so $x=\left(4\pm 6i\right)$x = (4 ± 6i) / $2=2\pm 3i$2 = 2 ± 3i. See Complex Roots of a Quadratic. The formula survives because completing the square, which produces it, never needs the square root to be real. The sign of the discriminant sorts the cases: a positive value gives two real roots, zero gives one repeated root, and only a negative value brings in i. Test the pair against the coefficients. The roots add to 2 + 2 = 4, the sum of their real parts, and multiply to $2^2+3^2=4+9=13$2² + 3² = 4 + 9 = 13, which match $-\frac{b}{a}$−b/a and $\frac{c}{a}$c/a for this equation.

The two roots are conjugates of each other. When the coefficients are real, non-real roots always come in conjugate pairs.

So a cubic with real coefficients has either three real roots or one real root and a conjugate pair, and its graph always crosses the x-axis. Given one non-real root, its conjugate is a second root. Multiply the two matching factors to get a real quadratic, then divide the cubic by it to find the real root. See Complex Roots of a Cubic and the quadratics and polynomials guide.

What does a complex number look like?

A complex number looks like a point on a plane. Plot a + bi at (a, b), with the real part along the horizontal axis and the imaginary part up the vertical axis; this plane is called the Argand diagram. See The Argand Diagram. The point for −2 + 3i sits 2 to the left of the origin and 3 up. A point can also be given by its distance from the origin, the modulus, and its angle from the positive real axis, the argument. A plane is needed because a complex number has two independent parts, and a line has room for only one. Multiplying by i turns every point a quarter turn counterclockwise: $i\left(2+i\right)=2i+i^2=-1+2i$i(2 + i) = 2i + i² = −1 + 2i, so (2, 1) moves to (−1, 2). Lengths confirm the turn. Both points are $\sqrt{5}$√5 from the origin, since $2^2+1^2=5$2² + 1² = 5 and ${\left(-1\right)}^2+2^2=1+4=5$(−1)² + 2² = 1 + 4 = 5.

The modulus |z| is its distance from the origin. The arrow to a + bi is the hypotenuse of a right-angled triangle with legs a and b, so $|z|=\sqrt{a^2+b^2}$|z| = √(a² + b²). For 3 + 4i the modulus is $\sqrt{9+16}=5$√(9 + 16) = 5. See The Modulus of a Complex Number.

The argument is the angle the arrow makes with the positive real axis, measured counterclockwise, and in the same triangle $\tan \theta =\frac{b}{a}$tan θ = b/a. A calculator's inverse tangent only returns angles between −90° and 90°, so for a point in the second or third quadrant, sketch it and adjust by 180°. See The Argument of a Complex Number.

Writing a number by its modulus r and argument $\theta$θ is polar form: $z=r\left(\cos \theta +i\sin \theta \right)$z = r(cos θ + i sin θ), because the real part is $r\cos \theta$r cos θ and the imaginary part is $r\sin \theta$r sin θ. See Polar Form and The Form $r\left(\cos \theta +i\sin \theta \right)$r(cos θ + i sin θ).

Division reverses both steps: divide the moduli and subtract the arguments. See Dividing in Polar Form.

What is De Moivre's theorem?

De Moivre's theorem is the rule for whole-number powers in polar form: $\left[r\left(\cos \theta +i\sin \theta \right)\right]$[r(cos θ + i sin θ)]ⁿ = rⁿ$\left(\cos n\theta +i\sin n\theta \right)$(cos nθ + i sin nθ). Raising z to the power n raises the modulus to the power n and multiplies the argument by n. For z = 2(cos 30° + i sin 30°), the cube has modulus 2 × 2 × 2 = 8 and argument 90°, three times 30°, so $z^3=8i$z³ = 8i. See De Moivre's Theorem. The theorem holds because multiplying complex numbers multiplies their moduli and adds their arguments, and a power is repeated multiplication; induction on n makes this a proof. Large powers need one extra step: an argument past 180° is brought back into the range −180° to 180° by subtracting 360°, so 300° becomes −60°. Confirm the example in Cartesian form: $z=\sqrt{3}+i$z = √3 + i, its square is $2+2\sqrt{3}i$2 + 2√3 i, and multiplying that by $\sqrt{3}+i$√3 + i gives 8i.

One use is to derive trigonometric identities. Expand ${\left(\cos \theta +i\sin \theta \right)}^3$(cos θ + i sin θ)³ with the binomial theorem and match the real part against $\cos 3\theta$cos 3θ. Replace ${\sin }^2\theta$sin²θ with $1-{\cos }^2\theta$1 − cos²θ to get $\cos 3\theta =4{\cos }^3\theta -3\cos \theta$cos 3θ = 4cos³θ − 3cos θ. See Multiple Angles by De Moivre.

The reverse direction turns a power of $\cos \theta$cos θ into cosines of multiple angles, the form needed before integrating. If |z| = 1, then $\frac{1}{z}$1/z is the conjugate of z, so $z+\frac{1}{z}=2\cos \theta$z + 1/z = 2 cos θ and zⁿ $+\frac{1}{z}$+ 1/zⁿ $=2\cos n\theta$= 2 cos nθ. See The Identity $z+\frac{1}{z}=2\cos \theta$z + 1/z = 2 cos θ.

De Moivre also sums series of cosines: the sum of $\cos k\theta$cos kθ is the real part of the sum of z^k, which is a geometric series with ratio z. See Summing a Series with De Moivre and the sequences and series guide.

The roots of unity

The equation zⁿ = 1 has n solutions, each of modulus 1, with arguments 0°, $\frac{360^{\circ }}{n}$360°/n, $2\times \frac{360^{\circ }}{n}$2 × 360°/n and so on, because each of those angles times n is a whole number of turns. See The Roots of Unity.

Joined in order, they are the vertices of a regular polygon: the fifth roots of 1 form a regular pentagon. See Regular Polygons from the Roots of Unity.

The same method finds the nth roots of any complex number. Take the real nth root of the modulus, divide the argument by n, then step around by $\frac{360^{\circ }}{n}$360°/n to collect the rest. The cube roots of 8i all have modulus 2, at 30°, 150° and 270°. See The nth Roots of a Complex Number.

Behind all of this is the exponential form. Substitute $x=i\theta$x = iθ into the series for e^x and the terms sort into the series for cosine and sine, so e^$i\theta$iθ $=\cos \theta +i\sin \theta$= cos θ + i sin θ and every complex number is re^$i\theta$iθ. De Moivre is then the index law (e^$i\theta$iθ)ⁿ = e^in$\theta$θ. See The Exponential Form of a Complex Number.

How do you read a complex equation as a picture?

Read every modulus as a distance and every argument as a direction: |z − a| is the distance from z to the point a, so |z − a| = r is the circle with center a and radius r. For example, |z − 3 + 2i| = 4 is the circle of radius 4 centered at 3 − 2i, because z − 3 + 2i is z − (3 − 2i). Always rewrite the expression as z minus one number before reading off the center. The picture works because z − a is the arrow from a to z, and a modulus is a length. With two moduli, |z − 1| = |z − 5| says z is as far from 1 as from 5, so it is the perpendicular bisector: the vertical line through (1 + 5) ÷ 2 = 3. Test a point: z = 7 − 2i makes z − 3 + 2i equal to 4, whose modulus is 4.

|z − a| = |z − b| says z is the same distance from a as from b, so z lies on the perpendicular bisector of the segment from a to b. See Circles and Bisectors on the Argand Diagram.

Fixing an argument gives a half-line: arg(z − a) = 45° is the ray from a pointing at 45°. Changing an equals sign to an inequality shades a region: $|z|\le 3$|z| ≤ 3 is the whole disc of radius 3, and a strict inequality leaves out the boundary. See Rays and Regions on the Argand Diagram.

The mistakes worth naming

• Losing the sign on $i^2$i². $-10i^2$−10i² is +10, because $i^2=-1$i² = −1.
• Trusting the calculator's inverse tangent for the argument. It returns an angle between −90° and 90°. Sketch the point, and add or subtract 180° in the second or third quadrant.
• Writing $\sqrt{-4}\times \sqrt{-9}=\sqrt{36}=6$√(−4) × √(−9) = √36 = 6. The rule $\sqrt{a}\times \sqrt{b}=\sqrt{ab}$√a × √b = √(ab) holds only for non-negative a and b. Convert first: $2i\times 3i=6i^2=-6$2i × 3i = 6i² = −6.
• Forgetting that non-real roots come in pairs. If 2 + 3i is a root and the coefficients are real, then 2 − 3i is also a root.
• Giving only one nth root. A complex number has three cube roots, evenly spaced around a circle.

Where this leads next

Rotation is multiplication by a complex number here and a 2 × 2 matrix in the matrices guide. The identity e^$i\theta$iθ $=\cos \theta +i\sin \theta$= cos θ + i sin θ comes from the Maclaurin series in the sequences and series guide, and every polynomial in the quadratics and polynomials guide has its full set of roots once complex numbers are allowed.