Why do we need complex numbers if they're imaginary?

Complex numbers are not 'imaginary' in the sense of fake or unreal; the term is historical. They are essential in engineering (AC circuits, signal processing), physics (quantum mechanics), and pure mathematics because i^2 = -1 allows us to solve equations that have no real solutions and work with a complete, closed number system. The imaginary unit i is as real and useful as any other number.

How do I multiply complex numbers without memorizing the formula?

Treat a + bi as a binomial and use the distributive property (FOIL): (a + bi)(c + di) = ac + adi + bci + bdi^2. Then substitute i^2 = -1: ac + adi + bci + bd(-1) = (ac - bd) + (ad + bc)i. This derivation is often easier to remember than the formula itself, so if you forget, re-derive it from first principles.

Can the imaginary part be negative, like in 3 - 2i?

Yes; the number 3 - 2i has real part 3 and imaginary part -2 (the coefficient of i is negative). In the formula (a + bi)(c + di), if b or d is negative, their signs are part of their values. The '+' in 'a + bi' is literal only when both parts are positive; always use the actual signed values when applying the multiplication formula.

What is the difference between complex numbers and imaginary numbers?

A complex number has the form a + bi where a and b are real numbers. If a \neq 0 and b \neq 0, it is complex with both real and imaginary parts. If a = 0 and b \neq 0, the number bi is called purely imaginary. The term 'imaginary numbers' sometimes refers only to purely imaginary numbers (like 3i), and sometimes to the entire complex number system; the exact meaning depends on context. On this page, 'complex numbers' is the formal term for all numbers of the form a + bi.

Complex Numbers

Solve equations with no real solutions and multiply combinations of real and imaginary numbers using the complex number system.

i^2 = -1, \quad (a + bi)(c + di) = (ac - bd) + (ad + bc)i

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What each symbol means

What Complex Numbers takes

Symbol	Meaning
$a$	The real part of the complex number $a + bi$ ; if $a = 0$ and $b \neq 0$ the number is purely imaginary.
$b$	The imaginary part (the coefficient of $i$ ) in $a + bi$ ; it is always a real number that multiplies the imaginary unit.
$c$	The real part of the second complex number $c + di$ in a multiplication or operation.
$d$	The imaginary part (the coefficient of $i$ ) in the second complex number $c + di$ .
$i$	The imaginary unit defined by $i^2 = -1$ ; it is not a variable but a fixed symbol representing $\sqrt{-1}$ .

When to use it

When you need to solve polynomial equations that have no real solutions or work with square roots of negative numbers.

Level

Usually taught in: Algebra II

Worked examples

1. Multiply two complex numbers with positive parts

Problem

Multiply

(2 + 3i)(1 + 4i)

$(2 + 3i)(1 + 4i) = (a+bi)(c+di) \text{ where } a=2, b=3, c=1, d=4$
Identify which numbers play which role in the formula $(a + bi)(c + di)$ .
$(ac - bd) + (ad + bc)i$
The formula tells us the real part is $ac - bd$ and the imaginary part is $ad + bc$ .
$ac = 2 \cdot 1 = 2, \quad bd = 3 \cdot 4 = 12$
Multiply the real parts and multiply the imaginary coefficients separately.
$ad = 2 \cdot 4 = 8, \quad bc = 3 \cdot 1 = 3$
These cross-products will form the imaginary part.
$(2 - 12) + (8 + 3)i = -10 + 11i$
The real part is $2 - 12 = -10$ ; the imaginary part is $8 + 3 = 11$ .

Answer: $-10 + 11i$

The formula $(a + bi)(c + di) = (ac - bd) + (ad + bc)i$ comes from distributing like you would with binomials, but the $i^2 = -1$ rule collapses the $bdi^2$ term into a real number. This example shows the straightforward case where all numbers are positive integers.

2. Multiply complex numbers with negative coefficients

Problem

Multiply

(-1 + 2i)(2 - 3i)

$(-1 + 2i)(2 - 3i) = (a+bi)(c+di) \text{ where } a=-1, b=2, c=2, d=-3$
Identify the coefficients, including their signs: $a = -1$ and $d = -3$ are negative.
$ac = (-1)(2) = -2, \quad bd = (2)(-3) = -6$
Calculate the products carefully: $ac$ is negative, and $bd$ is also negative (positive times negative).
$ac - bd = -2 - (-6) = -2 + 6 = 4$
Subtracting a negative is the same as adding: $-2 - (-6) = -2 + 6 = 4$ . This is the real part.
$ad = (-1)(-3) = 3, \quad bc = (2)(2) = 4$
Negative times negative gives positive for $ad$ ; the imaginary part coefficient comes from $ad + bc = 3 + 4$ .
$4 + (3 + 4)i = 4 + 7i$
Combine the real and imaginary parts: real is $4$ , imaginary coefficient is $7$ .

Answer: $4 + 7i$

When complex numbers have negative coefficients, the key is tracking signs carefully through the multiplications, especially in the $ac - bd$ term where subtracting a negative can flip the sign. This example shows that the formula handles all sign combinations correctly without special cases.

3. Complex impedance in an electrical circuit

Problem

A physicist measures impedances in an AC circuit. The first component has impedance

Z_1 = 3 + 2i

ohms (where the imaginary part represents reactance), and a second has impedance

Z_2 = 1 - i

ohms. Find the combined effect by computing

Z_1 \cdot Z_2

$Z_1 \cdot Z_2 = (3 + 2i)(1 - i)$
Set up the multiplication of impedances as complex numbers with $a=3, b=2, c=1, d=-1$ .
$ac = 3 \cdot 1 = 3, \quad bd = 2 \cdot (-1) = -2$
The real-part product is $ac = 3$ ; the imaginary-part product is $bd = -2$ (positive times negative).
$ac - bd = 3 - (-2) = 3 + 2 = 5$
The real part of the result is $3 - (-2) = 5$ . Subtracting the negative $bd$ increases the real part.
$ad = 3 \cdot (-1) = -3, \quad bc = 2 \cdot 1 = 2$
Cross-multiply to get the coefficient of $i$ : $ad + bc = -3 + 2 = -1$ .
$5 + (-1)i = 5 - i$
The result is a complex impedance with real part $5$ and imaginary coefficient $-1$ , written as $5 - i$ .

Answer: $5 - i$

In electrical engineering, impedance is a complex number because AC circuits involve both resistance (real part) and reactance (imaginary part). Multiplying impedances requires the complex multiplication formula, demonstrating how abstract algebra solves real-world problems in physics and engineering.

Common mistakes

Where Complex Numbers usually goes wrong

The mistake	Why it is wrong	The fix
Writing $(a + bi)(c + di) = ac + bd + adi + bci$ without applying $i^2 = -1$ .	This mistake comes from distributing the terms as if $i$ were just another variable, but $i^2 = -1$ is not a property of a variable—it defines the imaginary unit, and the $bdi^2$ term must be converted to $-bd$ (a real number).	Always substitute $i^2 = -1$ to turn $bdi^2$ into $-bd$ , giving the correct formula $(ac - bd) + (ad + bc)i$ .
After computing $(2 + 3i)(1 + 2i) = 2 + 4i + 3i + 6i^2 = 2 + 7i - 6 = -4 + 7i$ , then later writing $-4 + 7i^2$ or treating the $7i$ term as $(7i)^2 = 49i^2 = -49$ .	Once $i^2$ has been evaluated and replaced with $-1$ , the simplification is complete. Trying to apply exponent rules a second time, or to an expression that already contains the simplified form, adds layers of confusion and breaks the arithmetic.	After substituting $i^2 = -1$ once, stop; the result $-4 + 7i$ is final. Do not square $i$ or any multiple of $i$ again unless the problem explicitly asks you to multiply the result by itself.
Treating complex multiplication as componentwise addition: $(2 + 3i)(1 + 2i) = (2+1) + (3+2)i = 3 + 5i$ .	Complex number multiplication is not defined as adding the real parts and adding the imaginary parts separately. The full distributive property requires four pairwise products: $2 \cdot 1$ , $2 \cdot 2i$ , $3i \cdot 1$ , and $3i \cdot 2i$ , and the last one involves $i^2$ .	Always use the complete multiplication formula $(a + bi)(c + di) = (ac - bd) + (ad + bc)i$ , which accounts for all four products and the $i^2 = -1$ rule.

Tips and when to use something else

Remember that $i^2 = -1$ is the defining property; every complex multiplication reduces to this rule at some point.
The formula $(ac - bd) + (ad + bc)i$ has a useful structure: the real part uses subtraction ( $ac - bd$ ) and the imaginary part uses addition ( $ad + bc$ ), which comes directly from applying the distributive property and the $i^2 = -1$ rule.
If you're solving a polynomial like $x^2 + 1 = 0$ , complex numbers give you solutions where classical algebra says 'no solution': $x = \pm i$ . Without complex numbers, many polynomial equations remain unsolvable.
For division or simplifying fractions with complex denominators, the complex conjugate method ( $a + bi$ times $a - bi = a^2 + b^2$ ) is often faster than multiplication—look up 'Complex Conjugate' when you encounter those problems.

Related formulas

Distributive PropertyThe Distributive Property lets you multiply a number by a sum or difference by distributing the multiplication to each term separately.Polynomial Long DivisionPolynomial long division breaks a fraction into quotient and remainder, letting you simplify rational expressions or test polynomial divisibility.Fundamental Theorem of AlgebraEvery polynomial equation of degree n has exactly n complex roots (counting multiplicity) — this guarantees solutions always exist.

Frequently asked questions

Why do we need complex numbers if they're imaginary?: Complex numbers are not 'imaginary' in the sense of fake or unreal; the term is historical. They are essential in engineering (AC circuits, signal processing), physics (quantum mechanics), and pure mathematics because $i^2 = -1$ allows us to solve equations that have no real solutions and work with a complete, closed number system. The imaginary unit $i$ is as real and useful as any other number.
How do I multiply complex numbers without memorizing the formula?: Treat $a + bi$ as a binomial and use the distributive property (FOIL): $(a + bi)(c + di) = ac + adi + bci + bdi^2$ . Then substitute $i^2 = -1$ : $ac + adi + bci + bd(-1) = (ac - bd) + (ad + bc)i$ . This derivation is often easier to remember than the formula itself, so if you forget, re-derive it from first principles.
Can the imaginary part be negative, like in $3 - 2i$ ?: Yes; the number $3 - 2i$ has real part $3$ and imaginary part $-2$ (the coefficient of $i$ is negative). In the formula $(a + bi)(c + di)$ , if $b$ or $d$ is negative, their signs are part of their values. The ' $+$ ' in ' $a + bi$ ' is literal only when both parts are positive; always use the actual signed values when applying the multiplication formula.
What is the difference between complex numbers and imaginary numbers?: A complex number has the form $a + bi$ where $a$ and $b$ are real numbers. If $a \neq 0$ and $b \neq 0$ , it is complex with both real and imaginary parts. If $a = 0$ and $b \neq 0$ , the number $bi$ is called purely imaginary. The term 'imaginary numbers' sometimes refers only to purely imaginary numbers (like $3i$ ), and sometimes to the entire complex number system; the exact meaning depends on context. On this page, 'complex numbers' is the formal term for all numbers of the form $a + bi$ .

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Reviewed 2026-09-18