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=−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.
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+bi and z2=c+di:
The operation treats the real and imaginary components independently. No interaction occurs between them during addition — the real sum a+c ignores the imaginary values, and the imaginary sum b+d ignores the real values.
Example
Add 3+2i and 1−5i.
(3+2i)+(1−5i)=(3+1)+(2+(−5))i=4+(−3)i=4−3i
The real parts 3 and 1 sum to 4. The imaginary parts 2 and −5 sum to −3. The result is 4−3i.
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+z1) and associative ((z1+z2)+z3=z1+(z2+z3)). Order and grouping do not affect the result.
z₁ + z₂ as the diagonal of a parallelogram
The two arrows z₁ and z₂ are placed tail to tail and the sum is the diagonal of the parallelogram they span, exactly as vectors add: real parts add along the horizontal, imaginary parts along the vertical. Drag either summand and watch the diagonal follow on the complex addition and subtraction tool.
Componentwise addition and vector addition are one and the same rule.
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.
Each component undergoes its own subtraction independently.
Example
Compute (7+4i)−(2+6i).
(7+4i)−(2+6i)=(7−2)+(4−6)i=5+(−2)i=5−2i
The real parts 7−2=5. The imaginary parts 4−6=−2. The result is 5−2i.
A common error occurs when the second number has a negative imaginary part. Consider (5+3i)−(2−4i). The subtraction must distribute across both terms:
Subtraction of Complex Numbers
(5+3i)−(2−4i)=5+3i−2+4i=3+7i
Forgetting to negate the −4i term — leaving it as −4i instead of +4i — produces the wrong answer 3−i. Always distribute the negative sign before combining terms.
Geometrically, z1−z2 represents the vector from z2 to z1. Its modulus∣z1−z2∣ 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 z2 to reach z1?
z₁ − z₂ = z₁ + (−z₂)
The subtrahend is reversed into −z₂ and then added: the difference is the diagonal of the parallelogram spanned by z₁ and the reversed arrow. Its length is the distance between the two original points. Reverse either operand and watch the diagonal swing on the complex addition and subtraction tool.
Subtraction is addition of the additive inverse, on the page and on the plane.
Multiplication of Complex Numbers
Multiplication requires more care than addition. The distributive property expands the product into four terms, and the identity i2=−1 collapses one of those terms into the real part.
For z1=a+bi and z2=c+di, expand using distribution (the FOIL method):
Multiplication of Complex Numbers
(a+bi)(c+di)=ac+adi+bci+bdi2
The term bdi2 contains i2=−1, so bdi2=−bd. Collecting real and imaginary parts:
Multiplying by a pure real number scales both components proportionally. The product 3(2+5i)=6+15i stretches the number by factor 3 without rotating it. Multiplying by i rotates 90° counterclockwise: i(2+5i)=2i+5i2=−5+2i. The trigonometric form reveals multiplication's full geometric meaning — moduli multiply, arguments add.
Fact: Squaring builds Pythagorean triples
For whole numbers a>b>0, the parts of (a+bi)2 are the legs of a right triangle with hypotenuse a2+b2, since ∣z2∣=∣z∣2: hence 3,4,5 and 5,12,13. Every primitive Pythagorean triple arises from such a square.
Squaring is the special case z1=z2. For z=a+bi the formula gives z2=(a2−b2)+2abi: the real part is a difference of two squares, and the imaginary part is twice the product of the two parts. With whole numbers this gives (2+i)2=3+4i, (3+2i)2=5+12i and (4+i)2=15+8i, and the modulus of each square is the square of the original modulus: 5, 13 and 17.
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
z₁z₂: arguments add, moduli multiply
The product sits at the angle θ₁ + θ₂ marked in the picture and at distance |z₁||z₂| from the origin: multiplying by a complex number rotates by its argument and scales by its modulus. Nothing in the algebraic expansion shows this; the picture does. Turn one factor and watch the product swing round on the complex multiplication tool.
The four-term expansion hides a rotation; the trigonometric form reveals it.
Division of Complex Numbers
Division presents a challenge absent in the other operations: a complex denominator violates the standard algebraic forma+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+bi and z2=c+di with z2=0:
The denominator becomes c2+d2, a real number, because z⋅zˉ=∣z∣2 for any complex z. The numerator requires standard complex multiplication. Once the denominator is real, divide each component of the numerator by it.
Step-by-step example: Compute 1−4i3+2i.
Identify the conjugate of the denominator: 1−4i=1+4i.
Multiply numerator and denominator:
Division of Complex Numbers
1−4i3+2i⋅1+4i1+4i=(1−4i)(1+4i)(3+2i)(1+4i)
Compute the denominator: (1−4i)(1+4i)=1−16i2=1+16=17.
Compute the numerator: (3+2i)(1+4i)=3+12i+2i+8i2=3+14i−8=−5+14i.
Divide: 17−5+14i=−175+1714i.
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
z₁/z₂: arguments subtract, moduli divide
The quotient sits at the angle θ₁ − θ₂ and at distance |z₁|/|z₂|: division undoes the rotation and the scaling that multiplication performs. Multiplying numerator and denominator by the conjugate is the algebra that produces this picture. Move the divisor and watch the quotient counter-rotate on the complex division tool.
The conjugate trick is the algebra that performs this counter-rotation.
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:
Properties of Complex Arithmetic
z1+z2=z2+z1z1⋅z2=z2⋅z1
Order does not matter. Adding 3+2i to 1−5i yields the same result as adding 1−5i to 3+2i. The same holds for products.
Associativity allows regrouping without changing results:
When adding or multiplying three or more complex numbers, parentheses may be placed anywhere.
Distributivity connects addition and multiplication:
Properties of Complex Arithmetic
z1⋅(z2+z3)=z1⋅z2+z1⋅z3
Multiplication distributes over sums, enabling expansion of products and factoring of expressions.
Identity elements exist for both operations. The additive identity is 0=0+0i, satisfying z+0=z for every z. The multiplicative identity is 1=1+0i, satisfying z⋅1=z for every z.
Inverse elements also exist. Every complex number z has an additive inverse −z with z+(−z)=0. Every nonzero complex number has a multiplicative inverse z−1 with z⋅z−1=1. These properties make C a mathematical field.
The Multiplicative Inverse
THEOREM
Every nonzero complex number possesses a multiplicative inverse — a number that multiplies with it to produce 1.
Finding this inverse employs the same conjugate technique used for division.
For z=a+bi=0, the multiplicative inverse is:
The Multiplicative Inverse
z−1=z1=∣z∣2zˉ=a2+b2a−bi
The formula multiplies 1/z by zˉ/zˉ, converting the denominator to the real value ∣z∣2=a2+b2. The result expresses z−1 in standard algebraic form.
The inverse exists precisely when ∣z∣2=0, which occurs exactly when z=0. Division by zero remains undefined in C as in R — the number 0 has no multiplicative inverse.
Geometrically, z−1 lies on the same ray from the origin as zˉ, but at distance 1/∣z∣ rather than ∣z∣. Inversion reflects across the real axis and rescales the modulus to its reciprocal.
Fact: Reciprocals bend lines into circles
Inverting every number 1+ti on a vertical line gives points of one circle: 1+ti1−21=2(1+t2)(1−ti)2 always has modulus 21. In general, taking reciprocals sends lines and circles to lines and circles.
Inversion moves whole sets of points, not only single numbers. Take the vertical line of numbers 1+ti, with t running through all real values, and invert each one. The point 1 stays where it is, 1+i goes to 21−21i and 1−i goes to 21+21i, while points far up or far down the line land ever closer to 0.
Inverting a nonzero complex number is a reflection plus a rescaling: take the conjugate, then divide by ∣z∣2.
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 thati2=−1: The most frequent error in multiplication. After expanding (a+bi)(c+di), the term bdi2 must become −bd, not +bd. Every i2 in any calculation converts to −1 without exception.
Sign errors in subtraction: When subtracting (c+di), both the real part c and the imaginary part di must be negated. The expression (5+3i)−(2−4i) becomes 5+3i−2+4i, not 5+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 1+i3+2i is incomplete. Multiply by the conjugate of the denominator to obtain proper form: (1+i)(1−i)(3+2i)(1−i)=25−i=25−21i.
Division by zero: The expression z/0 is undefined for any z. No multiplicative inverse of 0 exists, and division by zero produces no meaningful result. Always verify denominators are nonzero before dividing.
Confusing conjugate and negative: The conjugate of 3+2i is 3−2i, not −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∣. For division, verify that multiplying the quotient by the divisor recovers the dividend.
Multiplying part by part
The product is formed the way a sum is: real part times real part, imaginary part times imaginary part. The two cross terms are lost, and the product of the imaginary parts lands in the wrong place.
Wrong working: (1+2i)(3+4i)=1⋅3+(2⋅4)i=3+8i.
Right working: (1+2i)(3+4i)=3+4i+6i+8i2=3+10i−8=−5+10i.
Why it happens: addition and subtraction really do work part by part, and the same pattern is carried over to multiplication. Multiplication distributes instead, so each part of one factor meets each part of the other.
How to check: moduli multiply. ∣1+2i∣⋅∣3+4i∣=5⋅5=125, and ∣−5+10i∣=25+100=125, while ∣3+8i∣=73.
Dividing part by part
The quotient is formed the way a difference is: real part over real part, imaginary part over imaginary part. Division by a complex number does not split over the parts of the denominator.
Wrong working: 2+3i4+6i=24+36i=2+2i.
Right working: multiply by the conjugate 2−3i. (2+3i)(2−3i)(4+6i)(2−3i)=4+98−12i+12i−18i2=1326=2.
Why it happens: addition and subtraction work part by part, and so does division by a real number: 24+6i=2+3i. A complex denominator mixes the parts, just as a complex factor does in multiplication.
How to check: multiply the answer by the denominator; it must give the numerator back. (2+2i)(2+3i)=4+6i+4i+6i2=−2+10i, not 4+6i, while 2(2+3i)=4+6i.
Stopping at the conjugate
The conjugate is found and given as the inverse, without the division by ∣z∣2. The answer points in the right direction but has the wrong length, unless ∣z∣=1.
Wrong working: (2−i)−1=2−i=2+i.
Right working: ∣2−i∣2=4+1=5, so (2−i)−1=52+i=52+51i. Indeed (2−i)(52+51i)=54+52i−52i−51i2=54+51=1.
Why it happens: the conjugate does the visible work of the formula, flipping the sign of the imaginary part, and the real divisor ∣z∣2 looks like a final tidy-up that can be dropped. On the unit circle the two coincide, which makes the shortcut look right.
How to check: multiply z by the answer; the product must be 1. Here (2−i)(2+i)=5, so the conjugate is 5 times too large.
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=−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.
Why do you multiply by the conjugate when dividing complex numbers?
+
Because a number times its conjugate gives a real result, which clears i out of the denominator. Multiplying numerator and denominator by the conjugate of the denominator leaves the value unchanged while turning the bottom into a plain real number. The result can then be split back into standard form. Try it on the division explorer.Read more →
Why isn't (a + bi)(c + di) just ac + bdi?
+
Because multiplication expands over all four pairs, not component by component. Every term multiplies every other, and the product of the two imaginary terms contains i squared, which collapses to −1 and moves into the real part. That extra contribution is exactly what makes complex multiplication a rotation combined with a stretch, not a stretch alone.Read more →
How do you square a complex number?
+
Expand as you would any binomial, then replace i squared with −1. For a + bi the result is a² − b² for the real part and 2ab for the imaginary part. Squaring in polar form is faster: square the modulus and double the argument. Explore both on the multiplication tool.Read more →
Do the ordinary rules of algebra apply to complex numbers?
+
Yes. Addition and multiplication remain commutative and associative, multiplication still distributes over addition, and every non-zero number still has a reciprocal. The one rule that does not carry over is ordering: complex numbers cannot be arranged as greater or smaller, so inequalities have no meaning between them.Read more →
Can you divide complex numbers by dividing the real and imaginary parts separately?
+
Only when the divisor is real: 24+6i=2+3i. A complex divisor mixes the parts, as a complex factor does in multiplication, so 2+3i4+6i is not 24+36i=2+2i. Since 4+6i=2(2+3i), the quotient is 2. Multiplying back by the divisor exposes the error: (2+2i)(2+3i)=−2+10i, not 4+6i.Read more →
How does squaring a complex number give Pythagorean triples?
+
For whole numbers a>b>0, the square (a+bi)2=(a2−b2)+2abi has parts that are the legs of a right triangle with hypotenuse a2+b2. The reason is ∣z2∣=∣z∣2: the length of the square is the square of the length. So (2+i)2=3+4i gives 3,4,5, and (3+2i)2=5+12i gives 5,12,13. Every primitive triple arises from such a square.Read more →
What happens to a straight line when you take reciprocals?
+
A line that misses the origin becomes a circle through it. Inverting every number 1+ti on the vertical line through 1 gives points of the circle of radius 21 centred at 21: the point 1 stays put, 1+i goes to 21−21i, and points far up or down the line land ever closer to 0. In general, taking reciprocals sends lines and circles to lines and circles.Read more →
