Arithmetic with Complex Numbers

Intuition

In the last lesson you met complex numbers, sums like $2 + 3i$ that pair a real part with an imaginary part. The natural next question is how to compute with them: how to add, subtract, multiply, and divide them. Each of those operations always lands back on a clean answer of the form $a + bi$ . The good news is that you already know almost every step. Treat $i$ like an ordinary variable, and collect the real terms with the real terms and the imaginary terms with the imaginary terms. The one moment anything new happens is when an $i^2$ appears, at which point you replace it with $-1$ . Division looks harder, but it turns out to be the same conjugate trick you used to rationalize a denominator. That trick is aimed now at clearing $i$ instead of a square root.

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 both parts by the denominator's 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 $\operatorname{Im}(z)$ . Notice that the imaginary part is the real coefficient $b$ , not $bi$ : in $2 + 3i$ the real part is $2$ and the imaginary part is $3$ .

Two familiar kinds of number are just special cases. When $b = 0$ the number is $a + 0i = a$ , an ordinary real number, so every real number is already a complex number. When $a = 0$ the number is $0 + bi = bi$ , a pure imaginary number like $3i$ or $-i$ . Standard form holds both of these, 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 $b = d$ . This is not an extra rule we impose; it follows from $i$ being non-real. If $a + bi = c + di$ with $b \ne d$ , then rearranging gives $(a - c) = (d - b)i$ , and dividing by the nonzero $d - b$ would make $i = \frac{a - c}{d - b}$ a ratio of real numbers, hence real. That contradicts $i^2 = -1$ , so the imaginary parts must be equal, and then the real parts must be equal too. Matching real and imaginary parts is therefore the only way two complex numbers can be equal, 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.

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

Write the sum out and reorder it using the commutative and associative laws of addition, which let you add real quantities in any order and grouping:

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

The two imaginary terms $bi$ and $di$ share the common factor $i$ , so the distributive law pulls it out:

bi + di = (b + d)i.

Putting the pieces back together gives $(a + c) + (b + d)i$ , which is again in standard form. In that form, the real part is the sum $a + c$ of the real parts, and the imaginary part is the sum $b + d$ of the imaginary parts. Nothing about $i^2$ ever came up, because no two factors of $i$ were multiplied together.

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 $(x + 2)(x + 3)$ . The one extra step is that the product of the two imaginary terms produces $i^2$ , which you immediately replace 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 $ad + bc$ . The $-bd$ in the real part is the whole reason a product of complex numbers is interesting. That $-bd$ is the $i^2 = -1$ substitution reaching back to change a term you might have expected to stay positive.

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

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.

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 pattern $(x + y)(x - y) = x^2 - y^2$ from the special-factorizations chapter, used here with $x = a$ and $y = bi$ :

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

The middle terms $-abi$ and $+abi$ are exact opposites and cancel, which is what clears the imaginary part. 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. The result 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.

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. Clearing it uses the same move you learned for rationalizing a radical denominator in the special-factorizations chapter. There you multiplied by the conjugate to turn $a + \sqrt{b}$ into the rational $a^2 - b$ ; here you multiply by the conjugate to turn $c + di$ into the real $c^2 + d^2$ . The imaginary unit plays the role the radical played, and the conjugate clears it the same way.

To divide $\dfrac{a + bi}{c + di}$ , 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 $c^2 + d^2$ , by the conjugate property just proved. Expanding the numerator and dividing each part by that real number puts the answer back in standard form:

\frac{a + bi}{c + di} = \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2} = \frac{ac + bd}{c^2 + d^2} + \frac{bc - ad}{c^2 + d^2}\,i.

As always with a conjugate, $\dfrac{c - di}{c - di} = 1$ , so multiplying by it changes the form of the quotient without changing its value. You never need the boxed formula; in practice you multiply by the conjugate, simplify the numerator with $i^2 = -1$ , and divide the real denominator into each part.

Worked example 4 Divide and write the answer in standard form

Divide $\dfrac{1 + 3i}{1 + i}$ . The denominator’s conjugate is $1 - i$ , and multiplying top and bottom by it makes the denominator $1^2 + 1^2 = 2$ :

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

Divide both parts of the numerator by $2$ :

\frac{4 + 2i}{2} = 2 + i.

The quotient comes out to the tidy $2 + i$ . A quick check 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 cancelled 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.

These four operations are the whole of complex arithmetic, and they are enough to give every quadratic equation a solution, the thread the next lessons pick up.

Common mistakes

Forgetting that $i^2 = -1$ . In a product, the $i^2$ term is neither zero nor left positive: in $(2 - i)(3 + 2i)$ the term $-2i^2$ becomes $+2$ and merges into the real part. Skipping this step is the single most frequent error.
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. $(6 - 2i) - (3 + 4i)$ is $3 - 6i$ , not $3 + 2i$ . The minus sign applies to both parts of the second number, the $3$ and the $4i$ .
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: $\dfrac{a + bi}{c + di} = \dfrac{(a + bi)(c - di)}{c^2 + d^2}$ .
Square with the binomial pattern: $(a + bi)^2 = (a^2 - b^2) + 2abi$ .

Arithmetic with Complex Numbers

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)

Free Response Questions (FRQ)