Arithmetic with Complex Numbers Advanced. This lesson goes beyond core Algebra I. You can skip it.

Learning goals

Write a complex number in standard form $a + bi$
Add and subtract by combining the parts separately
Multiply like binomials, replacing $i^2$ with $-1$
Use the conjugate, since $(a + bi)(a - bi)$ is real
Divide by multiplying the numerator and denominator by the conjugate
Square with $(a + bi)^2 = (a^2 - b^2) + 2abi$

The standard form a + bi

A complex number is a number written in the standard form

z = a + bi,

where $a$ and $b$ are real numbers and $i$ is the imaginary unit with $i^2 = -1$ . The real number $a$ is the real part of $z$ , written $\operatorname{Re}(z)$ , and the real number $b$ is the imaginary part, written $\operatorname{Im}(z)$ . The imaginary part is the coefficient $b$ , not the term $bi$ : in $2 + 3i$ the real part is $2$ and the imaginary part is $3$ . A real number is the case $b = 0$ , and a pure imaginary number like $3i$ is the case $a = 0$ ; standard form covers both, and everything between, in one shape.

Because the two parts play different roles, two complex numbers count as equal only when they match part for part: $a + bi = c + di$ exactly when $a = c$ and $b = d$ . That is not an extra rule bolted on; it follows from $i$ being non-real, a fact the next lessons lean on constantly.

Adding and subtracting

Addition and subtraction are the easiest operations, because the real and imaginary parts never interact. You simply combine them separately. For $(3 + i) + (1 + 2i)$ , that means adding $3 + 1 = 4$ and $1 + 2 = 3$ to get $4 + 3i$ .

Why $(a + bi) + (c + di) = (a + c) + (b + d)i$ #

Reorder the sum so the real terms sit together and the imaginary terms sit together:

(a + bi) + (c + di) = (a + c) + (bi + di).

The two imaginary terms $bi$ and $di$ share the factor $i$ , so they combine into $(b + d)i$ :

bi + di = (b + d)i.

Putting the pieces back together gives $(a + c) + (b + d)i$ , already in standard form. The real part is the sum of the real parts, and the imaginary part is the sum of the imaginary parts. No two factors of $i$ were ever multiplied together, so $i^2$ never comes up in addition.

So the rules are

\begin{aligned} (a + bi) + (c + di) &= (a + c) + (b + d)i, \\ (a + bi) - (c + di) &= (a - c) + (b - d)i. \end{aligned}

Subtraction works the same way once you distribute the minus sign to both parts of the second number. The single most common slip is to subtract the real parts but forget to subtract the imaginary parts. To avoid that slip, keep the whole second number inside its parentheses until the sign is distributed.

Worked example 1 Add and subtract two complex numbers

Add $(3 + 5i) + (4 - 2i)$ by combining the parts separately:

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

Now subtract $(7 + 2i) - (1 + 6i)$ . Distribute the minus sign to both the $1$ and the $6i$ :

(7 + 2i) - (1 + 6i) = (7 - 1) + (2 - 6)i = 6 - 4i.

Each answer is already in standard form, the real part first and a single imaginary term second.

There is a clean picture behind this. Suppose you plot a complex number $a + bi$ as the point $a$ units along a horizontal real axis and $b$ units up a vertical imaginary axis. Then adding two complex numbers adds their horizontal steps and adds their vertical steps, which is exactly how you add arrows tip to tail.

Plot a complex number a + bi by going a units along the real axis and b units up the imaginary axis. Adding combines the parts separately, so the sum lands at the far corner of the parallelogram built on the two arrows: (3 + i) + (1 + 2i) = 4 + 3i, because 3 + 1 = 4 and 1 + 2 = 3.

Check your understanding

Simplify $(6 - 2i) - (3 + 4i)$ .

3 - 6i

3 + 2i

9 - 6i

3 - 2i

Multiplying

To multiply two complex numbers, treat them as two binomials and expand with the distributive law, the same “first, outer, inner, last” you use for $(x + 2)(x + 3)$ . Try $(1 + i)(2 + i)$ : expanding gives $2 + i + 2i + i^2 = 2 + 3i - 1 = 1 + 3i$ . The one extra step is that the product of the two imaginary terms produces $i^2$ . That is the only new thing that ever happens, and you immediately replace it with $-1$ .

Why $(a + bi)(c + di) = (ac - bd) + (ad + bc)i$ #

Expand the product by multiplying every term of the first factor by every term of the second:

(a + bi)(c + di) = ac + adi + bci + bd\,i^2.

Three of these terms are ordinary products of real numbers times $1$ or $i$ . The last term carries $i^2$ , and this is the only place the imaginary unit does anything special. Replace $i^2$ with $-1$ :

bd\,i^2 = bd(-1) = -bd.

Now collect the real terms, $ac$ and $-bd$ , and the imaginary terms, $adi$ and $bci$ :

(a + bi)(c + di) = (ac - bd) + (ad + bc)i.

The real part is $ac - bd$ and the imaginary part is $ad + bc$ . That $-bd$ is the whole reason a product of complex numbers is interesting. It is where $i^2 = -1$ turns a term you might expect to stay positive into one that lowers the real part instead.

You do not need to memorize the final formula. It is faster and safer to expand each product by hand and replace $i^2$ with $-1$ as it appears, exactly as in the examples below. Two special cases are worth noticing: multiplying by a real number $k$ scales both parts, $k(a + bi) = ka + kbi$ , and multiplying by $i$ sends $a + bi$ to $i(a + bi) = ai + bi^2 = -b + ai$ . For example, $i(4 + 2i) = 4i + 2i^2 = -2 + 4i$ .

Worked example 2 Multiply two complex numbers

Expand $(2 + 3i)(4 - i)$ term by term:

(2 + 3i)(4 - i) = 8 - 2i + 12i - 3i^2.

Replace $i^2$ with $-1$ , so $-3i^2 = -3(-1) = +3$ , then combine like terms:

8 - 2i + 12i - 3i^2 = 8 + 10i + 3 = 11 + 10i.

Try a second, $(5 + 2i)(1 + 4i)$ :

(5 + 2i)(1 + 4i) = 5 + 20i + 2i + 8i^2 = 5 + 22i - 8 = -3 + 22i.

In both products the $i^2$ term flipped sign and merged into the real part, which is exactly where a beginner’s answer most often goes wrong.

Check your understanding

Multiply $(2 - i)(3 + 2i)$ .

8 + i

4 + i

8 - i

6 + i

The complex conjugate

Before dividing, we need one special product. The complex conjugate of $z = a + bi$ is the number with the sign of its imaginary part flipped,

\overline{z} = a - bi,

read “z bar.” Conjugation changes only the imaginary part: the conjugate of $2 + 3i$ is $2 - 3i$ , and the conjugate of $-4 - i$ is $-4 + i$ . What makes the conjugate useful is what happens when you multiply a number by it: try $(2 + i)(2 - i) = 4 - 2i + 2i - i^2 = 4 + 1 = 5$ , and every trace of $i$ is gone.

Why $(a + bi)(a - bi) = a^2 + b^2$ #

The two factors $a + bi$ and $a - bi$ are a sum and a difference of the same two terms, so their product is a difference of squares. That is the familiar pattern $(x + y)(x - y) = x^2 - y^2$ , applied here with $x = a$ and $y = bi$ :

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

Now expand the square $(bi)^2 = b^2 i^2$ and replace $i^2$ with $-1$ :

a^2 - (bi)^2 = a^2 - b^2 i^2 = a^2 - b^2(-1) = a^2 + b^2.

The difference of squares became a sum of squares, because $i^2 = -1$ turned the subtracted $b^2 i^2$ into an added $b^2$ . The result $a^2 + b^2$ is a real number, and since it is a sum of two squares it is never negative. It is zero only when $a$ and $b$ are both zero, that is, only when $z = 0$ .

That is the key property: a nonzero complex number times its conjugate is a positive real number, with every trace of $i$ gone. Multiplying $a + bi$ by $a - bi$ is the surest way to turn a complex number into a real one, and it is exactly the tool that makes division possible.

Worked example 3 Conjugates and their products

The conjugate of $4 - 3i$ is $4 + 3i$ . Their product clears the imaginary part:

(4 - 3i)(4 + 3i) = 4^2 + 3^2 = 16 + 9 = 25.

The conjugate of the pure imaginary number $5i$ is $-5i$ , and

(5i)(-5i) = -25i^2 = 25,

again a positive real number. In standard form $5i = 0 + 5i$ , so here $a = 0$ and $b = 5$ , and $a^2 + b^2 = 25$ as the pattern predicts.

Check your understanding

What is $(-4 - i)(-4 + i)$ ?

17

15

16 - i

-17

Dividing

A quotient of two complex numbers, such as $\dfrac{1 + 3i}{1 + i}$ , is not yet in standard form: it has an imaginary part sitting in the denominator. Multiply top and bottom by $1 - i$ , the denominator’s conjugate:

\frac{1 + 3i}{1 + i} = \frac{1 + 3i}{1 + i}\cdot\frac{1 - i}{1 - i} = \frac{(1 + 3i)(1 - i)}{(1 + i)(1 - i)} = \frac{4 + 2i}{2} = 2 + i.

Multiplying by $\dfrac{1 - i}{1 - i} = 1$ changes the fraction’s form without changing its value, the same conjugate move you used to rationalize a radical denominator. It works because $(1 + i)(1 - i) = 1^2 + 1^2 = 2$ , a real number, by the conjugate property just proved, so the $i$ in the denominator is gone.

That is the whole method. In general, to divide $\dfrac{a + bi}{c + di}$ (with $c + di \ne 0$ ), multiply the numerator and the denominator by the conjugate $c - di$ of the denominator:

\frac{a + bi}{c + di} = \frac{a + bi}{c + di}\cdot\frac{c - di}{c - di} = \frac{(a + bi)(c - di)}{c^2 + d^2}.

The denominator is now the real number $c^2 + d^2$ , never zero here since $c$ and $d$ are not both zero. From there, expand the numerator with $i^2 = -1$ and divide its real part and its imaginary part by that real denominator, exactly as above.

Worked example 4 Check a quotient, then divide a harder one

A quick check on the quotient above multiplies back: $(2 + i)(1 + i) = 2 + 2i + i + i^2 = 1 + 3i$ , the original numerator.

Not every quotient is that clean. Divide $\dfrac{2 + 3i}{3 + 2i}$ using the conjugate $3 - 2i$ , with denominator $3^2 + 2^2 = 13$ :

\frac{2 + 3i}{3 + 2i} = \frac{(2 + 3i)(3 - 2i)}{13} = \frac{6 - 4i + 9i - 6i^2}{13} = \frac{12 + 5i}{13}.

Split the single fraction into its real and imaginary parts to reach standard form:

\frac{12 + 5i}{13} = \frac{12}{13} + \frac{5}{13}i.

Fractions in the parts are perfectly normal; the answer is still $a + bi$ , now with $a = \frac{12}{13}$ and $b = \frac{5}{13}$ .

Check your understanding

Write $\dfrac{5 + i}{1 - i}$ in standard form $a + bi$ .

2 + 3i

3 + 2i

2 - 3i

4 + 6i

Powers of a complex number

Raising a complex number to a power is just repeated multiplication, so the same rules apply. The most common case is a square, which follows the square-of-a-binomial pattern $(x + y)^2 = x^2 + 2xy + y^2$ from earlier, again with a final $i^2 = -1$ :

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

The real part is $a^2 - b^2$ and the imaginary part is $2ab$ .

Worked example 5 Square a complex number

Square $2 + 3i$ with the binomial pattern, then apply $i^2 = -1$ :

(2 + 3i)^2 = 2^2 + 2(2)(3i) + (3i)^2 = 4 + 12i + 9i^2 = 4 + 12i - 9 = -5 + 12i.

A difference can collapse even further. Square $1 - i$ :

(1 - i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i.

Here the real parts canceled completely, leaving the pure imaginary $-2i$ . Higher powers build on the same idea: to cube a number, square it and multiply once more, reducing every $i^2$ as it appears.

Check your understanding

Square $3 + 2i$ .

5 + 12i

13 + 12i

5 + 6i

9 + 4i

Add, subtract, multiply, divide, and now square: with these tools, complex numbers are ready to use. The chapter now turns back to quadratics, to build a method that can reach the complex answers this arithmetic makes sense of.

Common mistakes

Forgetting that $i^2 = -1$ . A term such as $-3i^2$ becomes $+3$ , not $-3$ and not $0$ . Skipping this substitution, or getting its sign wrong, is the single most frequent error in a product.
Combining a real part with an imaginary part. $3 + 2i$ does not simplify to $5$ or to $5i$ . The two parts are different kinds of number and stay separate, just as $3 + 2x$ cannot be collapsed into a single term.
Dropping the minus sign in a subtraction. The minus sign applies to both parts of the second number, not just its real part. Distribute it before combining, and keep the second number inside its parentheses until you do.
Getting the conjugate product wrong. $(a + bi)(a - bi) = a^2 + b^2$ , a sum, not the $a^2 - b^2$ you would get with real numbers. The $i^2 = -1$ turns the subtracted square into an added one.
Conjugating the wrong part. The conjugate of $a + bi$ is $a - bi$ : only the imaginary part changes sign. It is not $-a + bi$ or $-a - bi$ .
Leaving $i$ in the denominator. A quotient is not finished until it is written as $a + bi$ . Multiply by the conjugate of the denominator, then divide each part by the resulting real number.

Key takeaways

A complex number in standard form is $a + bi$ with real part $a$ and imaginary part $b$ ; two are equal exactly when both parts match.
Add and subtract by combining the parts separately: $(a + bi) \pm (c + di) = (a \pm c) + (b \pm d)i$ .
Multiply by expanding like binomials and replacing $i^2$ with $-1$ : $(a + bi)(c + di) = (ac - bd) + (ad + bc)i$ .
The conjugate of $a + bi$ is $a - bi$ , and $(a + bi)(a - bi) = a^2 + b^2$ , a real, nonnegative number.
Divide by multiplying the numerator and denominator by the conjugate of the denominator, the same move as rationalizing a radical: $\dfrac{a + bi}{c + di} = \dfrac{(a + bi)(c - di)}{c^2 + d^2}$ , for a nonzero denominator $c + di$ .
Square with the binomial pattern: $(a + bi)^2 = (a^2 - b^2) + 2abi$ .

Arithmetic with Complex Numbers Advanced. This lesson goes beyond core Algebra I. You can skip it.

Learning goals

The standard form a + bi

Adding and subtracting

Multiplying

The complex conjugate

Dividing

Powers of a complex number

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice