Table of Contents
In this chapter we focus on how to do arithmetic with complex numbers that you already know how to write in the form
$$
a + bi,
$$
where $a$ and $b$ are real numbers and $i$ is the imaginary unit ($i^2 = -1$).
We will work with complex numbers written in this “standard” or rectangular form, and learn how to:
- add and subtract them
- multiply them
- find their conjugates
- divide them (by removing $i$ from the denominator using conjugates)
Throughout, $a,b,c,d$ will denote real numbers.
Equality of complex numbers
Two complex numbers are equal exactly when their real parts are equal and their imaginary parts are equal.
If
$$
a + bi = c + di,
$$
then
$$
a = c \quad\text{and}\quad b = d.
$$
This is often used to solve for unknowns: you expand both sides into the form (real part) $+$ (imaginary part)$\cdot i$ and then equate the coefficients of $1$ and $i$.
Example: Suppose
$$
(2x - 3) + (4y + 1)i = 5 + 7i.
$$
Equate real and imaginary parts:
- Real: $2x - 3 = 5 \implies 2x = 8 \implies x = 4$.
- Imaginary: $4y + 1 = 7 \implies 4y = 6 \implies y = \frac{3}{2}$.
Addition and subtraction
To add or subtract complex numbers in standard form, combine real parts with real parts and imaginary parts with imaginary parts.
Given
$$
(a + bi) + (c + di),
$$
you add:
$$
(a + bi) + (c + di) = (a + c) + (b + d)i.
$$
Similarly, for subtraction:
$$
(a + bi) - (c + di) = (a - c) + (b - d)i.
$$
You can think of this as just collecting like terms, with $1$ and $i$ playing the roles of “different kinds” of terms.
Examples
- Addition:
$$
(3 + 5i) + (2 - 7i)
= (3 + 2) + (5i - 7i)
= 5 - 2i.
$$ - Subtraction:
$$
(1 - 4i) - (6 + 3i)
= (1 - 6) + (-4i - 3i)
= -5 - 7i.
$$ - Adding more than two:
$$
(1 + 2i) + (-3 + 4i) + (5 - 6i)
= (1 - 3 + 5) + (2i + 4i - 6i)
= 3 + 0i = 3.
$$
Multiplication
To multiply complex numbers in standard form, distribute (use the same idea as expanding products of binomials), and then use $i^2 = -1$ to simplify.
Given
$$
(a + bi)(c + di),
$$
expand:
$$
(a + bi)(c + di)
= ac + adi + bci + bdi^2.
$$
Now replace $i^2$ by $-1$:
$$
= ac + adi + bci - bd
= (ac - bd) + (ad + bc)i.
$$
This is a useful formula to remember:
$$
(a + bi)(c + di) = (ac - bd) + (ad + bc)i.
$$
Examples
- Basic multiplication:
$$
(2 + 3i)(4 - i)
$$
Expand:
$$
(2 + 3i)(4 - i)
= 2\cdot 4 + 2\cdot(-i) + 3i\cdot 4 + 3i\cdot(-i)
= 8 - 2i + 12i - 3i^2.
$$
Since $i^2 = -1$,
$$
-3i^2 = -3(-1) = 3,
$$
so
$$
8 - 2i + 12i + 3
= 11 + 10i.
$$ - Multiplying by a real number (a special case):
$$
5(2 - 7i) = (5\cdot2) + (5\cdot(-7))i = 10 - 35i.
$$
This just scales both parts. - Squaring a complex number:
$$
(3 + 4i)^2 = (3 + 4i)(3 + 4i).
$$
Use the same method:
$$
(3 + 4i)(3 + 4i) = 3\cdot3 + 3\cdot4i + 4i\cdot3 + 4i\cdot4i
= 9 + 12i + 12i + 16i^2.
$$
With $i^2 = -1$,
$$
16i^2 = 16(-1) = -16,
$$
so
$$
9 + 24i - 16 = -7 + 24i.
$$ - Product of a number and its “negative”:
$$
(1 + 2i)(1 - 2i)
= 1\cdot1 + 1\cdot(-2i) + 2i\cdot1 + 2i\cdot(-2i)
= 1 - 2i + 2i - 4i^2.
$$
The $-2i + 2i$ cancel, leaving
$$
1 - 4i^2 = 1 - 4(-1) = 1 + 4 = 5.
$$
Notice that this last example turns a complex product into a purely real number. This is not an accident and is related to the idea of the conjugate, which we discuss next.
Complex conjugates
For a complex number $z = a + bi$, its complex conjugate is
$$
\overline{z} = a - bi.
$$
So, conjugation just flips the sign of the imaginary part and leaves the real part unchanged.
Some useful facts (you do not need to prove them here, just use them):
- $\overline{\overline{z}} = z$ (taking the conjugate twice gives you back $z$).
- If $z$ is real (i.e. $b = 0$), then $\overline{z} = z$.
- $z + \overline{z} = 2a$ is always real.
- $z - \overline{z} = 2bi$ is always purely imaginary.
Most importantly for our purposes:
$$
(a + bi)(a - bi) = a^2 + b^2.
$$
The product of a complex number and its conjugate is always a nonnegative real number.
Example
Let $z = 3 - 4i$. Then $\overline{z} = 3 + 4i$, and
$$
z\overline{z} = (3 - 4i)(3 + 4i)
= 3^2 + 4^2 = 9 + 16 = 25.
$$
This property is the key tool we use to divide complex numbers and remove $i$ from denominators.
Division of complex numbers
Division with complex numbers is defined in the usual algebraic way, but we almost always want to express the result back in standard form $x + yi$. To do that, we make sure the denominator is real by multiplying by a suitable form of $1$.
To divide
$$
\frac{a + bi}{c + di},
$$
with $c + di \neq 0$, multiply numerator and denominator by the conjugate of the denominator, $c - di$:
$$
\frac{a + bi}{c + di} \cdot \frac{c - di}{c - di}.
$$
The denominator becomes a real number:
$$
(c + di)(c - di) = c^2 + d^2.
$$
Then expand the numerator and simplify.
In general:
$$
\frac{a + bi}{c + di}
= \frac{(a + bi)(c - di)}{(c + di)(c - di)}
= \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2}.
$$
If you like, you can then split it into real and imaginary parts by dividing both parts by $c^2 + d^2$.
Examples
- Simple division:
$$
\frac{3 + 4i}{1 - 2i}.
$$
Multiply by the conjugate of the denominator, + 2i$:
$$
\frac{3 + 4i}{1 - 2i} \cdot \frac{1 + 2i}{1 + 2i}
= \frac{(3 + 4i)(1 + 2i)}{(1 - 2i)(1 + 2i)}.
$$
Denominator:
$$
(1 - 2i)(1 + 2i) = 1^2 + 2^2 = 5.
$$
Numerator:
$$
(3 + 4i)(1 + 2i)
= 3\cdot1 + 3\cdot2i + 4i\cdot1 + 4i\cdot2i
= 3 + 6i + 4i + 8i^2
= 3 + 10i - 8
= -5 + 10i.
$$
So
$$
\frac{3 + 4i}{1 - 2i} = \frac{-5 + 10i}{5} = -1 + 2i.
$$
- Division by a purely imaginary number:
$$
\frac{6 - 2i}{3i}.
$$
There are two common approaches.
One is to treat the denominator as $0 + 3i$ and multiply by its conjugate $0 - 3i = -3i$:
$$
\frac{6 - 2i}{3i} \cdot \frac{-3i}{-3i}
= \frac{(6 - 2i)(-3i)}{(3i)(-3i)}.
$$
Denominator:
$$
(3i)(-3i) = -9i^2 = -9(-1) = 9.
$$
Numerator:
$$
(6 - 2i)(-3i) = 6(-3i) + (-2i)(-3i) = -18i + 6i^2 = -18i - 6.
$$
So
$$
\frac{6 - 2i}{3i} = \frac{-6 - 18i}{9} = -\frac{2}{3} - 2i.
$$
Another approach is to write $3i = 3\cdot i$ and divide real and imaginary parts by $3$, then handle the division by $i$. For example, note that
$$
\frac{1}{i} = -i
$$
(because $i(-i) = -i^2 = 1$).
Both methods must give the same answer.
- Expressing a complex fraction in standard form:
$$
\frac{5 - i}{2 + 3i}.
$$
Multiply by the conjugate - 3i$:
$$
\frac{5 - i}{2 + 3i} \cdot \frac{2 - 3i}{2 - 3i}
= \frac{(5 - i)(2 - 3i)}{(2 + 3i)(2 - 3i)}.
$$
Denominator:
$$
(2 + 3i)(2 - 3i) = 2^2 + 3^2 = 4 + 9 = 13.
$$
Numerator:
$$
(5 - i)(2 - 3i)
= 5\cdot2 + 5(-3i) + (-i)\cdot2 + (-i)(-3i)
= 10 - 15i - 2i + 3i^2
= 10 - 17i + 3(-1)
= 7 - 17i.
$$
So
$$
\frac{5 - i}{2 + 3i} = \frac{7 - 17i}{13}
= \frac{7}{13} - \frac{17}{13}i.
$$
Combining operations
Once you know how to add, subtract, multiply, conjugate, and divide complex numbers, you can tackle expressions that mix several operations. The rules are the same as for real numbers:
- Follow the usual order of operations: parentheses, exponents, multiplication/division, addition/subtraction.
- At each step, simplify and rewrite in standard form $a + bi$.
Example
Simplify and write in standard form:
$$
(1 - 3i)^2 + \frac{4 + 2i}{1 + i}.
$$
- Square $(1 - 3i)$:
$$
(1 - 3i)^2 = (1 - 3i)(1 - 3i)
= 1 - 3i - 3i + 9i^2
= 1 - 6i + 9(-1)
= -8 - 6i.
$$ - Simplify the fraction:
$$
\frac{4 + 2i}{1 + i} \cdot \frac{1 - i}{1 - i}
= \frac{(4 + 2i)(1 - i)}{(1 + i)(1 - i)}.
$$
Denominator:
$$
(1 + i)(1 - i) = 1^2 + 1^2 = 2.
$$
Numerator:
$$
(4 + 2i)(1 - i)
= 4\cdot1 + 4(-i) + 2i\cdot1 + 2i(-i)
= 4 - 4i + 2i - 2i^2
= 4 - 2i + 2
= 6 - 2i.
$$
So
$$
\frac{4 + 2i}{1 + i} = \frac{6 - 2i}{2} = 3 - i.
$$
- Add the results:
$$
(-8 - 6i) + (3 - i)
= (-8 + 3) + (-6i - i)
= -5 - 7i.
$$
So the simplified answer is $-5 - 7i$.
Summary of main rules
For complex numbers $a + bi$ and $c + di$:
- Addition:
$$
(a + bi) + (c + di) = (a + c) + (b + d)i.
$$ - Subtraction:
$$
(a + bi) - (c + di) = (a - c) + (b - d)i.
$$ - Multiplication:
$$
(a + bi)(c + di) = (ac - bd) + (ad + bc)i.
$$ - Conjugate:
$$
\overline{a + bi} = a - bi, \quad (a + bi)(a - bi) = a^2 + b^2.
$$ - Division (with $c + di \neq 0$):
$$
\frac{a + bi}{c + di}
= \frac{(a + bi)(c - di)}{c^2 + d^2}
= \frac{ac + bd}{c^2 + d^2}
+ \frac{bc - ad}{c^2 + d^2}i.
$$
Every time you perform operations with complex numbers, your goal is to return to the standard form $x + yi$, with $x$ and $y$ real, using these rules.