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Operations on Complex Numbers



Key Terms
Addition of Complex Numbers
Subtraction of Complex Numbers
Multiplication of Complex Numbers
Division of Complex Numbers
Properties of Complex Arithmetic
The Multiplicative Inverse
Common Pitfalls and Tips
Summary: The Four Operations Side by Side



Computing with Two-Part Numbers

Complex numbers combine real and imaginary components, and arithmetic must handle both parts systematically. Addition and subtraction operate component-wise, just like vector arithmetic. Multiplication requires the distributive property and the fundamental identity i2=−1i^2 = -1. Division introduces the conjugate as an essential tool for eliminating imaginary denominators. These four operations extend real arithmetic into the complex plane while preserving the algebraic structure that makes calculation predictable.

Key Terms

Complex Number— the objects being operated on
Algebraic Form— the form used for addition and subtraction
Complex Conjugate— essential for division
Multiplicative Inverse— the reciprocal enabling division

See All Complex Numbers Definitions →


Addition of Complex Numbers

Adding complex numbers works exactly as intuition suggests: combine like terms. Real parts add to real parts, imaginary parts add to imaginary parts, and the two sums form the new complex number.

The formula states the rule precisely. For z1=a+biz_1 = a + bi and z2=c+diz_2 = c + di:

Addition
(a+bi)+(c+di)=(a+c)+(b+d)i(a + bi) + (c + di) = (a + c) + (b + d)i
Learn more about this formula: Addition →


The operation treats the real and imaginary components independently. No interaction occurs between them during addition — the real sum a+ca + c ignores the imaginary values, and the imaginary sum b+db + d ignores the real values.

Example: Add 3+2i3 + 2i and 1−5i1 - 5i.

(3+2i)+(1−5i)=(3+1)+(2+(−5))i=4+(−3)i=4−3i(3 + 2i) + (1 - 5i) = (3 + 1) + (2 + (-5))i = 4 + (-3)i = 4 - 3i


The real parts 33 and 11 sum to 44. The imaginary parts 22 and −5-5 sum to −3-3. The result is 4−3i4 - 3i.

On the complex plane, addition corresponds to vector addition. Place the two complex numbers as arrows from the origin. Position the second arrow so its tail sits at the head of the first. The sum extends from the origin to the final head — the familiar parallelogram or tip-to-tail rule from physics and geometry. This geometric interpretation makes addition visual and connects complex arithmetic to vector mechanics.

Addition is both commutative (z1+z2=z2+z1z_1 + z_2 = z_2 + z_1) and associative ((z1+z2)+z3=z1+(z2+z3)(z_1 + z_2) + z_3 = z_1 + (z_2 + z_3)). Order and grouping do not affect the result.

Subtraction of Complex Numbers

Subtraction mirrors addition with signs reversed. Subtract real parts separately, subtract imaginary parts separately, and combine the differences into the result.

The formula: for z1=a+biz_1 = a + bi and z2=c+diz_2 = c + di:

Subtraction
(a+bi)−(c+di)=(a−c)+(b−d)i(a + bi) - (c + di) = (a - c) + (b - d)i
Learn more about this formula: Subtraction →


Each component undergoes its own subtraction independently.

Example: Compute (7+4i)−(2+6i)(7 + 4i) - (2 + 6i).

(7+4i)−(2+6i)=(7−2)+(4−6)i=5+(−2)i=5−2i(7 + 4i) - (2 + 6i) = (7 - 2) + (4 - 6)i = 5 + (-2)i = 5 - 2i


The real parts 7−2=57 - 2 = 5. The imaginary parts 4−6=−24 - 6 = -2. The result is 5−2i5 - 2i.

A common error occurs when the second number has a negative imaginary part. Consider (5+3i)−(2−4i)(5 + 3i) - (2 - 4i). The subtraction must distribute across both terms:

(5+3i)−(2−4i)=5+3i−2+4i=3+7i(5 + 3i) - (2 - 4i) = 5 + 3i - 2 + 4i = 3 + 7i


Forgetting to negate the −4i-4i term — leaving it as −4i-4i instead of +4i+4i — produces the wrong answer 3−i3 - i. Always distribute the negative sign before combining terms.

Geometrically, z1−z2z_1 - z_2 represents the vector from z2z_2 to z1z_1. Its modulus ∣z1−z2∣|z_1 - z_2| gives the distance between the two points in the complex plane. Subtraction answers the question: how far and in what direction must one travel from z2z_2 to reach z1z_1?

Multiplication of Complex Numbers

Multiplication requires more care than addition. The distributive property expands the product into four terms, and the identity i2=−1i^2 = -1 collapses one of those terms into the real part.

For z1=a+biz_1 = a + bi and z2=c+diz_2 = c + di, expand using distribution (the FOIL method):

(a+bi)(c+di)=ac+adi+bci+bdi2(a + bi)(c + di) = ac + adi + bci + bdi^2


The term bdi2bdi^2 contains i2=−1i^2 = -1, so bdi2=−bdbdi^2 = -bd. Collecting real and imaginary parts:

Multiplication
(a+bi)(c+di)=(ac−bd)+(ad+bc)i(a + bi)(c + di) = (ac - bd) + (ad + bc)i
Learn more about this formula: Multiplication →


The real part of the product is ac−bdac - bd. The imaginary part is ad+bcad + bc. Both involve contributions from all four original values.

Step-by-step example: Compute (2+3i)(4−i)(2 + 3i)(4 - i).

Expand: 2⋅4+2⋅(−i)+3i⋅4+3i⋅(−i)=8−2i+12i−3i22 \cdot 4 + 2 \cdot (-i) + 3i \cdot 4 + 3i \cdot (-i) = 8 - 2i + 12i - 3i^2

Substitute i2=−1i^2 = -1: =8−2i+12i−3(−1)=8−2i+12i+3= 8 - 2i + 12i - 3(-1) = 8 - 2i + 12i + 3

Combine: =(8+3)+(−2+12)i=11+10i= (8 + 3) + (-2 + 12)i = 11 + 10i

Multiplying by a pure real number scales both components proportionally. The product 3(2+5i)=6+15i3(2 + 5i) = 6 + 15i stretches the number by factor 33 without rotating it. Multiplying by ii rotates 90°90° counterclockwise: i(2+5i)=2i+5i2=−5+2ii(2 + 5i) = 2i + 5i^2 = -5 + 2i. The trigonometric form reveals multiplication's full geometric meaning — moduli multiply, arguments add.
Step Action Worked example: (2 + 3i)(4 − i)
1 Expand using distributivity (FOIL) 8 − 2i + 12i − 3i²
2 Substitute i² = −1 8 − 2i + 12i + 3
3 Combine real and imaginary terms 11 + 10i

Division of Complex Numbers

Division presents a challenge absent in the other operations: a complex denominator violates the standard algebraic form a+bia + bi. The solution employs the conjugate to convert the denominator to a real number.

The technique: multiply both numerator and denominator by the conjugate of the denominator. For z1=a+biz_1 = a + bi and z2=c+diz_2 = c + di with z2≠0z_2 \neq 0:

Division
a+bic+di=(ac+bd)+(bc−ad)ic2+d2\frac{a + bi}{c + di} = \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2}
Learn more about this formula: Division →


The denominator becomes c2+d2c^2 + d^2, a real number, because z⋅zˉ=∣z∣2z \cdot \bar{z} = |z|^2 for any complex zz. The numerator requires standard complex multiplication. Once the denominator is real, divide each component of the numerator by it.

Step-by-step example: Compute 3+2i1−4i\frac{3 + 2i}{1 - 4i}.

Identify the conjugate of the denominator: 1−4i‾=1+4i\overline{1 - 4i} = 1 + 4i.

Multiply numerator and denominator:

3+2i1−4i⋅1+4i1+4i=(3+2i)(1+4i)(1−4i)(1+4i)\frac{3 + 2i}{1 - 4i} \cdot \frac{1 + 4i}{1 + 4i} = \frac{(3 + 2i)(1 + 4i)}{(1 - 4i)(1 + 4i)}


Compute the denominator: (1−4i)(1+4i)=1−16i2=1+16=17(1 - 4i)(1 + 4i) = 1 - 16i^2 = 1 + 16 = 17.

Compute the numerator: (3+2i)(1+4i)=3+12i+2i+8i2=3+14i−8=−5+14i(3 + 2i)(1 + 4i) = 3 + 12i + 2i + 8i^2 = 3 + 14i - 8 = -5 + 14i.

Divide: −5+14i17=−517+1417i\frac{-5 + 14i}{17} = -\frac{5}{17} + \frac{14}{17}i.

The result sits in standard form with real denominator eliminated.
Step Action Worked example: (3 + 2i) / (1 − 4i)
1 Identify the conjugate of the denominator 1 + 4i
2 Multiply numerator and denominator by that conjugate (3 + 2i)(1 + 4i) / [(1 − 4i)(1 + 4i)]
3 Simplify the denominator using z · z̄ = |z|² (3 + 2i)(1 + 4i) / 17
4 Expand and simplify the numerator (−5 + 14i) / 17
5 Write in standard form a + bi −5⁄17 + (14⁄17)i

Properties of Complex Arithmetic

Complex arithmetic inherits the structural properties that govern real number operations. These properties ensure that algebraic manipulation proceeds predictably, following rules familiar from elementary mathematics.

Commutativity holds for both addition and multiplication:

z1+z2=z2+z1z1⋅z2=z2⋅z1z_1 + z_2 = z_2 + z_1 \qquad z_1 \cdot z_2 = z_2 \cdot z_1


Order does not matter. Adding 3+2i3 + 2i to 1−5i1 - 5i yields the same result as adding 1−5i1 - 5i to 3+2i3 + 2i. The same holds for products.

Associativity allows regrouping without changing results:

(z1+z2)+z3=z1+(z2+z3)(z1⋅z2)⋅z3=z1⋅(z2⋅z3)(z_1 + z_2) + z_3 = z_1 + (z_2 + z_3) \qquad (z_1 \cdot z_2) \cdot z_3 = z_1 \cdot (z_2 \cdot z_3)


When adding or multiplying three or more complex numbers, parentheses may be placed anywhere.

Distributivity connects addition and multiplication:

z1⋅(z2+z3)=z1⋅z2+z1⋅z3z_1 \cdot (z_2 + z_3) = z_1 \cdot z_2 + z_1 \cdot z_3


Multiplication distributes over sums, enabling expansion of products and factoring of expressions.

Identity elements exist for both operations. The additive identity is 0=0+0i0 = 0 + 0i, satisfying z+0=zz + 0 = z for every zz. The multiplicative identity is 1=1+0i1 = 1 + 0i, satisfying z⋅1=zz \cdot 1 = z for every zz.

Inverse elements also exist. Every complex number zz has an additive inverse −z-z with z+(−z)=0z + (-z) = 0. Every nonzero complex number has a multiplicative inverse z−1z^{-1} with z⋅z−1=1z \cdot z^{-1} = 1. These properties make C\mathbb{C} a mathematical field.

The Multiplicative Inverse

Every nonzero complex number possesses a multiplicative inverse — a number that multiplies with it to produce 11. Finding this inverse employs the same conjugate technique used for division.

For z=a+bi≠0z = a + bi \neq 0, the multiplicative inverse is:

z−1=1z=zˉ∣z∣2=a−bia2+b2z^{-1} = \frac{1}{z} = \frac{\bar{z}}{|z|^2} = \frac{a - bi}{a^2 + b^2}


The formula multiplies 1/z1/z by zˉ/zˉ\bar{z}/\bar{z}, converting the denominator to the real value ∣z∣2=a2+b2|z|^2 = a^2 + b^2. The result expresses z−1z^{-1} in standard algebraic form.

Example: Find the multiplicative inverse of 3+4i3 + 4i.

Compute the modulus squared: ∣3+4i∣2=9+16=25|3 + 4i|^2 = 9 + 16 = 25.

Find the conjugate: 3+4i‾=3−4i\overline{3 + 4i} = 3 - 4i.

Apply the formula: (3+4i)−1=3−4i25=325−425i(3 + 4i)^{-1} = \frac{3 - 4i}{25} = \frac{3}{25} - \frac{4}{25}i.

Verification: (3+4i)⋅(325−425i)=925−1225i+1225i−1625i2=925+1625=2525=1(3 + 4i) \cdot \left(\frac{3}{25} - \frac{4}{25}i\right) = \frac{9}{25} - \frac{12}{25}i + \frac{12}{25}i - \frac{16}{25}i^2 = \frac{9}{25} + \frac{16}{25} = \frac{25}{25} = 1.

The inverse exists precisely when ∣z∣2≠0|z|^2 \neq 0, which occurs exactly when z≠0z \neq 0. Division by zero remains undefined in C\mathbb{C} as in R\mathbb{R} — the number 00 has no multiplicative inverse.

Geometrically, z−1z^{-1} lies on the same ray from the origin as zˉ\bar{z}, but at distance 1/∣z∣1/|z| rather than ∣z∣|z|. Inversion reflects across the real axis and rescales the modulus to its reciprocal.

Common Pitfalls and Tips

Complex arithmetic follows clear rules, but several common errors trap students repeatedly. Recognizing these pitfalls prevents mistakes and builds computational confidence.

Forgetting $i^2 = -1$: The most frequent error in multiplication. After expanding (a+bi)(c+di)(a + bi)(c + di), the term bdi2bdi^2 must become −bd-bd, not +bd+bd. Every i2i^2 in any calculation converts to −1-1 without exception.

Sign errors in subtraction: When subtracting (c+di)(c + di), both the real part cc and the imaginary part didi must be negated. The expression (5+3i)−(2−4i)(5 + 3i) - (2 - 4i) becomes 5+3i−2+4i5 + 3i - 2 + 4i, not 5+3i−2−4i5 + 3i - 2 - 4i. Distribute the negative sign completely before combining terms.

Leaving complex denominators: Final answers in algebraic form require real denominators. A result like 3+2i1+i\frac{3 + 2i}{1 + i} is incomplete. Multiply by the conjugate of the denominator to obtain proper form: (3+2i)(1−i)(1+i)(1−i)=5−i2=52−12i\frac{(3 + 2i)(1 - i)}{(1 + i)(1 - i)} = \frac{5 - i}{2} = \frac{5}{2} - \frac{1}{2}i.

Division by zero: The expression z/0z/0 is undefined for any zz. No multiplicative inverse of 00 exists, and division by zero produces no meaningful result. Always verify denominators are nonzero before dividing.

Confusing conjugate and negative: The conjugate of 3+2i3 + 2i is 3−2i3 - 2i, not −3−2i-3 - 2i. Conjugation flips only the imaginary sign; negation flips both. These operations differ and serve different purposes.

Tips for accuracy: Work step by step, writing each intermediate result. Check products by verifying that the modulus of the product equals the product of moduli: ∣z1z2∣=∣z1∣∣z2∣|z_1 z_2| = |z_1||z_2|. For division, verify that multiplying the quotient by the divisor recovers the dividend.
Pitfall Wrong Correct Why
Forgetting i² = −1 (2i)(3i) = 6i² = 6 (2i)(3i) = 6i² = −6 every i² must be replaced with −1
Sign error in subtraction (5 + 3i) − (2 − 4i) = 3 − i (5 + 3i) − (2 − 4i) = 3 + 7i distribute the minus across both real and imaginary parts
Leaving a complex denominator (3 + 2i) ⁄ (1 + i) as final answer multiply by conjugate → 5⁄2 − (1⁄2)i standard form a + bi requires a real denominator
Confusing conjugate with negative conjugate of 3 + 2i is −3 − 2i conjugate of 3 + 2i is 3 − 2i conjugation flips the imaginary sign only; negation flips both
Dividing by zero z ⁄ 0 treated as some value z ⁄ 0 is undefined 0 has no multiplicative inverse in ℂ (or ℝ)

Summary: The Four Operations Side by Side

The four operations on complex numbers share a common pattern — combine like terms for addition and subtraction, expand and apply i2=−1i^2 = -1 for multiplication, and convert to real-denominator form via the conjugate for division. The table below collects each operation's general formula, the key technique that makes it work, and a worked example side by side for quick reference.
Operation General formula Key technique Example
Addition (a + bi) + (c + di) = (a + c) + (b + d)i combine like terms component-wise (3 + 2i) + (1 − 5i) = 4 − 3i
Subtraction (a + bi) − (c + di) = (a − c) + (b − d)i distribute the minus, then combine (7 + 4i) − (2 + 6i) = 5 − 2i
Multiplication (a + bi)(c + di) = (ac − bd) + (ad + bc)i FOIL, then apply i² = −1 (2 + 3i)(4 − i) = 11 + 10i
Division (a + bi) ⁄ (c + di) = [(a + bi)(c − di)] ⁄ (c² + d²) multiply num and denom by conjugate of denom (3 + 2i) ⁄ (1 − 4i) = −5⁄17 + (14⁄17)i