In short

Two complex numbers are equal exactly when their real parts match and their imaginary parts match. Addition and subtraction work component-wise — add the real parts, add the imaginary parts. Multiplication uses the distributive law plus the rule i^2 = -1. The conjugate \bar{z} of z = a + bi is a - bi — flip the sign of the imaginary part. The conjugate turns division into something concrete: multiply top and bottom by the conjugate of the denominator, and the denominator becomes a real number.

Suppose you need to simplify (3 + 2i)(1 - 4i). You can FOIL it out like any binomial product -- but then one of the terms gives you i^2, and that is -1, so it folds back into the real part. The imaginary unit is not just a symbol you carry around; it actively reshapes the arithmetic. Addition, subtraction, multiplication, and division of complex numbers each have clean rules, and every one of them rests on a single fact: i^2 = -1.

Equality of complex numbers

Two complex numbers z_1 = a + bi and z_2 = c + di are equal if and only if

a = c \quad \text{and} \quad b = d.

Both the real parts and the imaginary parts must match independently. A single complex equation is equivalent to two real equations. This principle turns many complex-number problems into systems of simultaneous equations in real variables.

For instance, if z = x + yi and z = 5 - 3i, then x = 5 and y = -3. No further work needed — just read off the parts.

A more interesting case: suppose (2x - y) + (x + 3y)i = 7 + 11i. Matching real parts gives 2x - y = 7. Matching imaginary parts gives x + 3y = 11. Solve the system: from the first equation, y = 2x - 7; substitute into the second to get x + 3(2x - 7) = 11, so 7x = 32, giving x = 32/7 and y = 2(32/7) - 7 = 15/7.

Addition and subtraction

Adding two complex numbers is straightforward: add the real parts and add the imaginary parts separately.

Addition and subtraction of complex numbers

If z_1 = a + bi and z_2 = c + di, then

z_1 + z_2 = (a + c) + (b + d)i
z_1 - z_2 = (a - c) + (b - d)i

This is exactly what the distributive law would give you if you treated i as an ordinary algebraic symbol:

(a + bi) + (c + di) = (a + c) + (b + d)i.

The real and imaginary parts do not interact during addition. They are independent channels. Think of it as adding two-dimensional vectors: the x-components add, the y-components add, and neither affects the other.

Some quick examples:

(3 + 2i) + (1 + 4i) = 4 + 6i
(5 - i) + (-2 + 3i) = 3 + 2i
(7 + 2i) - (3 + 5i) = 4 - 3i
Addition of two complex numbers as vector addition on the complex planeA coordinate plane with the real axis horizontal and the imaginary axis vertical. Two arrows start from the origin: one to the point (3, 2) representing 3 plus 2i, and one to (1, 4) representing 1 plus 4i. A parallelogram is completed with dashed lines, and the diagonal from the origin reaches (4, 6), representing the sum 4 plus 6i. All three points are marked with red dots.ReIm012341234563 + 2i1 + 4i4 + 6i
Adding $3 + 2i$ and $1 + 4i$ on the complex plane. Each complex number is an arrow from the origin. The sum $4 + 6i$ is the diagonal of the parallelogram formed by the two arrows — exactly the same parallelogram rule you see in physics for adding forces or displacements.

Properties of addition

Addition of complex numbers inherits every structural property from real addition:

All of this follows because the real and imaginary parts each obey the rules for real numbers, and the two channels don't interact during addition.

Multiplication

Multiplication is where i^2 = -1 actually does something. Multiply (a + bi)(c + di) the same way you multiply any two binomials — distribute each term of the first across the second:

(a + bi)(c + di) = ac + adi + bci + bdi^2

Now use i^2 = -1:

= ac + adi + bci + bd(-1) = (ac - bd) + (ad + bc)i

Multiplication of complex numbers

If z_1 = a + bi and z_2 = c + di, then

z_1 \cdot z_2 = (ac - bd) + (ad + bc)i

You do not need to memorise this formula. Just expand the product as a binomial and replace i^2 with -1. The formula falls out every time.

Some examples:

(2 + 3i)(4 + i) = 8 + 2i + 12i + 3i^2 = 8 + 14i + 3(-1) = 5 + 14i
(1 + i)(1 - i) = 1 - i + i - i^2 = 1 - (-1) = 2

That second one is striking. The product of 1 + i and 1 - i is a pure real number — no imaginary part at all. The two factors are related by a sign flip on the imaginary part, and that sign flip is exactly what kills the imaginary component in the product. This leads to the next and most important idea.

Properties of multiplication

The complex numbers, with these properties, form a field — the same structure the reals have. Every rule from Operations and Properties that holds for real arithmetic also holds for complex arithmetic.

The conjugate

The complex conjugate of z = a + bi is written \bar{z} (or sometimes z^*) and is defined by

Complex conjugate

If z = a + bi, then

\bar{z} = a - bi.

The conjugate flips the sign of the imaginary part and leaves the real part unchanged.

The conjugate is the mirror image of z across the real axis. If z sits above the axis (positive imaginary part), \bar{z} sits the same distance below. If z is on the real axis, then \bar{z} = z — real numbers are their own conjugates.

A complex number and its conjugate as mirror images across the real axisA complex plane with the real axis horizontal and imaginary axis vertical. A red dot at (3, 2) is labelled z equals 3 plus 2i. Its mirror image at (3, negative 2) is labelled z-bar equals 3 minus 2i. A dashed vertical line connects them, crossing the real axis at (3, 0). The real axis is marked as the mirror line.ReIm012342i−2ii−iz = 3 + 2iz̄ = 3 − 2imidpoint on Re axis
The complex number $z = 3 + 2i$ and its conjugate $\bar{z} = 3 - 2i$ are reflections of each other across the real axis. The dashed line connecting them is perpendicular to the axis, and the midpoint lies on the axis at $3$. Every complex number and its conjugate share this mirror relationship.

Properties of the conjugate

The conjugate obeys a clean set of rules. Each one is easy to verify by writing z = a + bi and w = c + di and computing both sides.

  1. Conjugate of a conjugate. \overline{\bar{z}} = z. Flipping the sign twice gives back the original.

  2. Conjugate of a sum. \overline{z + w} = \bar{z} + \bar{w}. The conjugate distributes over addition.

  3. Conjugate of a product. \overline{z \cdot w} = \bar{z} \cdot \bar{w}. The conjugate distributes over multiplication.

  4. Product with own conjugate. z \cdot \bar{z} = a^2 + b^2. This is always a non-negative real number. It equals zero only when z = 0.

Property 4 is the most important. Write it out:

(a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - b^2 i^2 = a^2 + b^2.

The imaginary parts cancel perfectly, and you are left with the sum of squares of the real and imaginary parts. This quantity a^2 + b^2 is the square of the modulus of z — you will meet it formally in Complex Numbers — Modulus and Argument.

  1. Extracting parts from the conjugate.
\operatorname{Re}(z) = \frac{z + \bar{z}}{2}, \qquad \operatorname{Im}(z) = \frac{z - \bar{z}}{2i}.

Check: z + \bar{z} = (a + bi) + (a - bi) = 2a, so dividing by 2 gives a. And z - \bar{z} = 2bi, so dividing by 2i gives b.

Division

To divide complex numbers, you need to write \dfrac{z_1}{z_2} in the standard form a + bi. The key idea: multiply top and bottom by the conjugate of the denominator.

\frac{z_1}{z_2} = \frac{z_1}{z_2} \cdot \frac{\bar{z_2}}{\bar{z_2}} = \frac{z_1 \cdot \bar{z_2}}{z_2 \cdot \bar{z_2}}

By property 4 of the conjugate, the denominator z_2 \cdot \bar{z_2} is a real number (the sum of squares of the real and imaginary parts of z_2). So after the multiplication, the denominator is real, and you can split the fraction into real and imaginary parts by dividing each through.

This is the same move as rationalising a denominator in surd arithmetic — multiplying by the conjugate to clear the irrational part from the bottom. The technique is identical; only the "irrational" thing you are clearing is i instead of \sqrt{2}.

Quick example: compute \dfrac{3 + 2i}{1 - i}.

Multiply top and bottom by \overline{1 - i} = 1 + i:

\frac{(3 + 2i)(1 + i)}{(1 - i)(1 + i)} = \frac{3 + 3i + 2i + 2i^2}{1 - i^2} = \frac{3 + 5i - 2}{1 + 1} = \frac{1 + 5i}{2} = \frac{1}{2} + \frac{5}{2}i.

The denominator became 2 — pure real, clean, done.

An interactive explorer

The figure below lets you see how the product of two complex numbers behaves. Drag the two red points z_1 and z_2 around the complex plane. The readouts show the real and imaginary parts of each, and the value of their product z_1 \cdot z_2. Try placing z_2 at i (on the imaginary axis at height 1) and dragging z_1 around — you will see the product rotate z_1 by 90° counterclockwise.

Interactive complex multiplication explorerA coordinate plane representing the complex plane with the real axis horizontal and imaginary axis vertical, ranging from negative 5 to 5 on both axes. Two red draggable points represent complex numbers z1 and z2. A readout panel at the top shows the values of z1, z2, and their product z1 times z2 computed in real time. The reader can drag either point to explore how complex multiplication works.ReIm1234−1−2−3−4↔ drag either red point
Drag $z_1$ and $z_2$ around the complex plane. The readouts update to show each number and their product. Place $z_2$ at $i$ (the point $(0, 1)$) and move $z_1$ — the product always rotates $z_1$ by $90°$ counterclockwise. Place $z_2$ at $-1$ and the product is $-z_1$, a $180°$ rotation. Complex multiplication is rotation and scaling.

Worked examples

Example 1: Compute $(3 - 4i)(2 + 5i) + \overline{(1 + 2i)}$

This combines multiplication and conjugation. Handle each piece in turn.

Step 1. Multiply (3 - 4i)(2 + 5i).

= 3 \cdot 2 + 3 \cdot 5i + (-4i) \cdot 2 + (-4i)(5i)
= 6 + 15i - 8i - 20i^2
= 6 + 7i - 20(-1) = 6 + 7i + 20 = 26 + 7i

Why: expand as a binomial product, collect real and imaginary parts, and replace i^2 with -1. The product (-4i)(5i) = -20i^2 = +20 flips sign because of i^2 = -1 — that is where the "cross term" contributes to the real part.

Step 2. Find \overline{(1 + 2i)}.

\overline{1 + 2i} = 1 - 2i

Why: the conjugate flips only the sign of the imaginary part.

Step 3. Add the results.

(26 + 7i) + (1 - 2i) = 27 + 5i

Why: addition is component-wise. Real parts: 26 + 1 = 27. Imaginary parts: 7 + (-2) = 5.

Result: (3 - 4i)(2 + 5i) + \overline{(1 + 2i)} = 27 + 5i.

Step-by-step computation shown as a flowchart from multiplication through conjugation to the final sumThree computation boxes arranged vertically with arrows between them. The first box shows the multiplication (3 minus 4i)(2 plus 5i) equals 26 plus 7i. The second box shows the conjugate of 1 plus 2i equals 1 minus 2i. The third box shows the sum 26 plus 7i plus 1 minus 2i equals 27 plus 5i, outlined in red as the final answer.(3 − 4i)(2 + 5i) = 26 + 7iconj(1 + 2i) = 1 − 2iadd27 + 5iCheck: Re = 27, Im = 5Both real numbers ✓
The computation flows in two parallel branches — multiplication on the left, conjugation on the right — and merges in the addition step. The final answer $27 + 5i$ has real part $27$ and imaginary part $5$, both real numbers as expected.

Each operation — multiply, conjugate, add — follows a single mechanical rule. There is no guessing. The multiplication uses distributivity plus i^2 = -1. The conjugate flips the imaginary sign. The addition puts the channels back together.

Example 2: Express $\dfrac{5 + i}{2 + 3i}$ in the form $a + bi$

Division requires clearing the complex denominator by multiplying by its conjugate.

Step 1. Identify the conjugate of the denominator.

\overline{2 + 3i} = 2 - 3i

Why: the conjugate of 2 + 3i flips the sign of the imaginary part. Multiplying the denominator by its conjugate will produce a real number.

Step 2. Multiply numerator and denominator by 2 - 3i.

\frac{5 + i}{2 + 3i} \cdot \frac{2 - 3i}{2 - 3i}

Why: multiplying by \dfrac{2 - 3i}{2 - 3i} = 1 does not change the value of the expression. It is the complex-number version of rationalising the denominator.

Step 3. Expand the numerator.

(5 + i)(2 - 3i) = 10 - 15i + 2i - 3i^2 = 10 - 13i - 3(-1) = 13 - 13i

Why: standard binomial expansion. The i^2 = -1 term contributes +3 to the real part.

Step 4. Expand the denominator.

(2 + 3i)(2 - 3i) = 4 - 9i^2 = 4 + 9 = 13

Why: this is the product z \bar{z} = a^2 + b^2 — the difference-of-squares pattern, with i^2 = -1 turning the minus into a plus. The result is always real and positive (for non-zero z).

Step 5. Divide.

\frac{13 - 13i}{13} = 1 - i

Result: \dfrac{5 + i}{2 + 3i} = 1 - i.

Division of 5 plus i by 2 plus 3i on the complex plane showing the quotient at 1 minus iA complex plane with three points marked. The point 5 plus i is in the first quadrant at coordinates (5, 1). The point 2 plus 3i is at (2, 3). The quotient 1 minus i is at (1, negative 1) in the fourth quadrant. Dashed lines from the origin to each point indicate their positions. The quotient point is highlighted with a larger red dot and label.ReIm012345i−i2i−2i5 + i2 + 3i1 − i= (5+i) ÷ (2+3i)
Three points on the complex plane: the dividend $5 + i$, the divisor $2 + 3i$, and the quotient $1 - i$. Division "undoes" the multiplication — if you multiply $1 - i$ by $2 + 3i$, you get $(1 - i)(2 + 3i) = 2 + 3i - 2i - 3i^2 = 2 + i + 3 = 5 + i$, recovering the dividend exactly.

You can verify the answer by multiplying back: (1 - i)(2 + 3i) = 2 + 3i - 2i - 3i^2 = 2 + i + 3 = 5 + i. The quotient times the divisor gives the dividend, confirming the division.

Common confusions

Going deeper

If you can add, subtract, multiply, and divide complex numbers, and you understand what the conjugate does and why it makes division work, you have the full arithmetic toolkit. The material below is for readers who want to see the algebraic structure more clearly and where it connects to geometry.

The complex numbers form a field

In Operations and Properties, you saw that the real numbers satisfy closure, commutativity, associativity, distributivity, and the existence of identity and inverse elements — the package called a field. The complex numbers satisfy all the same properties. Closure is clear (the sum, difference, product, or quotient of two complex numbers is always a complex number). The structural properties all follow from the corresponding properties of the reals, because complex arithmetic is built on top of real arithmetic.

The one non-trivial piece is the multiplicative inverse. Given z = a + bi \neq 0, its inverse is

z^{-1} = \frac{\bar{z}}{z \bar{z}} = \frac{a - bi}{a^2 + b^2}.

This is well-defined because a^2 + b^2 > 0 whenever z \neq 0 (at least one of a, b is non-zero, so the sum of their squares is positive). You can verify: z \cdot z^{-1} = \dfrac{z \bar{z}}{a^2 + b^2} = \dfrac{a^2 + b^2}{a^2 + b^2} = 1.

So \mathbb{C} is a field. But unlike \mathbb{R}, it is not an ordered field — there is no way to define "<" on \mathbb{C} that is compatible with the field operations. You cannot say i > 0 or i < 0 without creating a contradiction. The trade-off for gaining algebraic closure (every polynomial splits into linear factors) is losing the ability to compare sizes. For size comparisons, you use the modulus |z|, which is always a real number.

Conjugation is an automorphism

The conjugation map z \mapsto \bar{z} respects both addition and multiplication: \overline{z + w} = \bar{z} + \bar{w} and \overline{z \cdot w} = \bar{z} \cdot \bar{w}. In the language of algebra, it is a field automorphism — a structure-preserving bijection from \mathbb{C} to itself. It is also an involution: applying it twice gives back the original (\overline{\bar{z}} = z).

The deeper significance: conjugation is the only non-trivial field automorphism of \mathbb{C} that fixes every real number. It is the unique symmetry of the complex numbers that leaves the reals unchanged. When you see complex roots of real polynomials appearing in conjugate pairs, it is because conjugation maps one root to the other while leaving the polynomial's real coefficients fixed.

Connection to matrices

There is a concrete way to represent complex numbers using 2 \times 2 real matrices. The complex number a + bi corresponds to the matrix

\begin{pmatrix} a & -b \\ b & a \end{pmatrix}.

Under this correspondence, matrix addition matches complex addition, and matrix multiplication matches complex multiplication. The conjugate a - bi corresponds to the transpose of the matrix. The modulus squared a^2 + b^2 is the determinant. This representation makes it concrete that complex arithmetic is nothing exotic — it is a specific pattern of real arithmetic packaged into a 2 \times 2 grid.

Where this leads next

The arithmetic on this page is the foundation for everything else in complex number theory.