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
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
This is exactly what the distributive law would give you if you treated i as an ordinary algebraic symbol:
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:
Properties of addition
Addition of complex numbers inherits every structural property from real addition:
- Commutative. z_1 + z_2 = z_2 + z_1.
- Associative. (z_1 + z_2) + z_3 = z_1 + (z_2 + z_3).
- Additive identity. z + 0 = z, where 0 = 0 + 0i.
- Additive inverse. The inverse of a + bi is -a - bi, since (a + bi) + (-a - bi) = 0.
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:
Now use i^2 = -1:
Multiplication of complex numbers
If z_1 = a + bi and z_2 = c + di, then
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:
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
- Commutative. z_1 \cdot z_2 = z_2 \cdot z_1.
- Associative. (z_1 \cdot z_2) \cdot z_3 = z_1 \cdot (z_2 \cdot z_3).
- Multiplicative identity. z \cdot 1 = z, where 1 = 1 + 0i.
- Distributive over addition. z_1(z_2 + z_3) = z_1 z_2 + z_1 z_3.
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
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.
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.
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Conjugate of a conjugate. \overline{\bar{z}} = z. Flipping the sign twice gives back the original.
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Conjugate of a sum. \overline{z + w} = \bar{z} + \bar{w}. The conjugate distributes over addition.
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Conjugate of a product. \overline{z \cdot w} = \bar{z} \cdot \bar{w}. The conjugate distributes over multiplication.
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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:
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.
- Extracting parts from the conjugate.
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.
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:
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.
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).
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)}.
Why: the conjugate flips only the sign of the imaginary part.
Step 3. Add the results.
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.
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.
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.
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.
Why: standard binomial expansion. The i^2 = -1 term contributes +3 to the real part.
Step 4. Expand the denominator.
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.
Result: \dfrac{5 + i}{2 + 3i} = 1 - i.
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
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"(a + bi)(c + di) = ac + bdi^2 — just multiply the corresponding parts." No. Multiplication is not component-wise. You must expand as a full binomial product and collect terms. The cross terms adi and bci contribute to the imaginary part of the product, and the bdi^2 term contributes to the real part (with a sign flip). Skipping the cross terms gives the wrong answer.
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"The conjugate of 3 + 2i is -3 - 2i." No — that is the negative, not the conjugate. The conjugate flips only the imaginary sign: \overline{3 + 2i} = 3 - 2i. The real part stays the same.
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"z \cdot \bar{z} is a complex number." It is always real. z \bar{z} = a^2 + b^2, which has no imaginary part. This is the key property that makes division work — multiplying the denominator by its conjugate produces a real denominator.
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"To divide, just divide the real parts and divide the imaginary parts." Division is not component-wise. You must multiply top and bottom by the conjugate of the denominator. There is no shortcut that avoids this step.
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"|z|^2 = z^2." No. |z|^2 = z \bar{z} = a^2 + b^2, while z^2 = (a + bi)^2 = a^2 - b^2 + 2abi, which is generally complex. The modulus squared uses the conjugate, not the square.
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"If z is complex, z + \bar{z} is complex." No — z + \bar{z} = 2a, which is always real. Similarly, z - \bar{z} = 2bi, which is always purely imaginary. These identities are often useful for extracting the real and imaginary parts.
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
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
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.
- Complex Numbers — Introduction — what i is and why it was needed, if you want to revisit the motivation.
- Operations and Properties — the field properties that \mathbb{C} inherits from \mathbb{R}, with the one new twist that ordering is lost.
- Complex Numbers — Modulus and Argument — the geometric meaning of |z| = \sqrt{a^2 + b^2} and the angle \theta that z makes with the positive real axis.
- Complex Numbers — Argand Plane — the full two-dimensional picture, where multiplication is rotation-and-scaling and conjugation is reflection.
- Quadratic Equations — Introduction — the quadratic formula that produces complex roots when D < 0, now fully interpreted.