# Multiplying complex numbers

Learn how to multiply two complex numbers. For example, multiply (1+2i)⋅(3+i).
A complex number is any number that can be written as start color greenD, a, end color greenD, plus, start color blueD, b, end color blueD, i, where i is the imaginary unit and start color greenD, a, end color greenD and start color blueD, b, end color blueD are real numbers.
When multiplying complex numbers, it's useful to remember that the properties we use when performing arithmetic with real numbers work similarly for complex numbers.
Sometimes, thinking of i as a variable, like x, is helpful. Then, with just a few adjustments at the end, we can multiply just as we'd expect. Let's take a closer look at this by walking through several examples.

## Multiplying a real number by a complex number

### Example

Multiply minus, 4, left parenthesis, 13, plus, 5, i, right parenthesis. Write the resulting number in the form of a, plus, b, i.

### Solution

If your instinct tells you to distribute the minus, 4, your instinct would be right! Let's do that!
\begin{aligned}\tealD{-4}(13+5i)&=\tealD{-4}(13)+\tealD{(-4)}(5i)\\ \\ &=-52-20i \end{aligned}
And that's it! We used the distributive property to multiply a real number by a complex number. Let's try something a little more complicated.

## Multiplying a pure imaginary number by a complex number

### Example

Multiply 2, i, left parenthesis, 3, minus, 8, i, right parenthesis. Write the resulting number in the form of a, plus, b, i.

### Solution

Again, let's start by distributing the 2, i to each term in the parentheses.
\begin{aligned}\tealD{2i}(3-8i)&=\tealD{2i}(3)-\tealD{2i}(8i)\\ \\ &=6i-16i^2 \end{aligned}
At this point, the answer is not of the form a, plus, b, i since it contains i, start superscript, 2, end superscript.
However, we know that start color goldD, i, start superscript, 2, end superscript, equals, minus, 1, end color goldD. Let's substitute and see where that gets us.
\begin{aligned}\phantom{\tealD{2i}(3-8i)} &=6i-16\goldD{i^2}\\ \\ &=6i-16(\goldD{-1})\\ \\ &=6i+16\\ \end{aligned}
Using the commutative property, we can write the answer as 16, plus, 6, i, and so we have that 2, i, left parenthesis, 3, minus, 8, i, right parenthesis, equals, 16, plus, 6, i.

### Problem 1

Multiply 3, left parenthesis, minus, 2, plus, 10, i, right parenthesis.

To multiply the two numbers, just distribute the 3 and simplify. This is shown below.
\begin{aligned}\tealD{3}(-2+10i)&=\tealD{3}(-2)+\tealD{(3)}(10i)\\ \\ &=-6+30i \end{aligned}

### Problem 2

Multiply minus, 6, i, left parenthesis, 5, plus, 7, i, right parenthesis.

Let's start by distributing the minus, 6, i to each term in the parentheses.
\begin{aligned}\tealD{-6i}(5+7i)&=\tealD{-6i}(5)+(\tealD{-6i})(7i)\\ \\ &=-30i+(-42i^2) \end{aligned}
Since start color goldD, i, start superscript, 2, end superscript, equals, minus, 1, end color goldD, this becomes:
\begin{aligned}\phantom{\tealD{-6i}(5+7i)} &=-30i+(-42\goldD{i^2})\\ \\ &=-30i+(-42(\goldD{-1}))\\ \\ &=-30i+42\\ \\ &=42-30i \end{aligned}
Excellent! We're now ready to step it up even more! What follows is the more typical case that you'll see when you're asked to multiply complex numbers.

## Multiplying two complex numbers

### Example

Multiply left parenthesis, 1, plus, 4, i, right parenthesis, left parenthesis, 5, plus, i, right parenthesis. Write the resulting number in the form of a, plus, b, i.

### Solution

In this example, some find it very helpful to think of i as a variable.
In fact, the process of multiplying these two complex numbers is very similar to multiplying two binomials! Multiply each term in the first number by each term in the second number.
\begin{aligned}(\tealD{1}+\maroonD{4i}) (5+i)&=(\tealD{1})(5)+(\tealD{1})(i)+(\maroonD{4i})(5)+(\maroonD{4i})(i)\\ \\ &=5+i+20i+4i^2\\ \\ &=5+21i+4i^2 \end{aligned}
Since start color goldD, i, start superscript, 2, end superscript, equals, minus, 1, end color goldD, we can replace i, start superscript, 2, end superscript with minus, 1 to obtain the desired form of a, plus, b, i.
\begin{aligned}\phantom{(\tealD{1}\maroonD{-5}i) (-6+i)} &=5+21i+4\goldD{i^2}\\ \\ &=5+21i+4(\goldD{-1})\\ \\ &=5+21i-4\\ \\ &=1+21i \end{aligned}

### Problem 3

Multiply left parenthesis, 1, plus, 2, i, right parenthesis, left parenthesis, 3, plus, i, right parenthesis.

Start by multiplying each term in the first number by each term in the second number.
\begin{aligned}(1+2i) (3+i)&={1}(3)+1(i)+{2i}(3)+{2i}(i)\\ \\ &=3+i+6i+2i^2\\ \\ &=3+7i+2i^2 \end{aligned}
Since start color goldD, i, start superscript, 2, end superscript, equals, minus, 1, end color goldD, we can replace i, start superscript, 2, end superscript with minus, 1 to obtain the desired form of a, plus, b, i.
\begin{aligned}\phantom{(\tealD{1}\maroonD{-5}i) (-6+i)} &=3+7i+2\goldD{i^2}\\ \\ &=3+7i+2(\goldD{-1})\\ \\ &=3+7i-2\\ \\ &=1+7i \end{aligned}

### Problem 4

Multiply left parenthesis, 4, plus, i, right parenthesis, left parenthesis, 7, minus, 3, i, right parenthesis.

Start by multiplying each term in the first number by each term in the second number.
\begin{aligned}(4+i) (7-3i)&=4(7)+4(-3i)+{i}(7)+{i}(-3i)\\ \\ &=28-12i+7i+(-3i^2)\\ \\ &=28-5i-3i^2 \end{aligned}
Since start color goldD, i, start superscript, 2, end superscript, equals, minus, 1, end color goldD, we can replace i, start superscript, 2, end superscript with minus, 1 to obtain the desired form of a, plus, b, i.
\begin{aligned}\phantom{(\tealD{1}\maroonD{-5}i) (-6+i)} &=28-5i-3\goldD{i^2}\\ \\ &=28-5i-3(\goldD{-1})\\ \\ &=28-5i+3\\ \\ &=31-5i \end{aligned}

### Problem 5

Multiply left parenthesis, 2, minus, i, right parenthesis, left parenthesis, 2, plus, i, right parenthesis.

\begin{aligned}(2-i) (2+i)&=2(2)+2(i)+({-i})(2)+({-i})(i)&\small{\gray{\text{Multiply}}}\\ \\ &=4+2i-2i-i^2&\small{\gray{\text{Simplify}}}\\ \\ &=4-i^2&\small{\gray{\text{Simplify}}}\\ \\ &=4-(-1)&\small{\gray{i^2=-1}}\\ \\ &=5&\small{\gray{\text{Simplify}}} \end{aligned}

### Problem 6

Multiply left parenthesis, 1, plus, i, right parenthesis, left parenthesis, 1, plus, i, right parenthesis.

\begin{aligned}(1+i)(1+i) &=1(1)+1(i)+i(1)+i(i)&\small{\gray{\text{Multiply}}}\\ \\ &=1+i+i+i^2&\small{\gray{\text{Simplify}}}\\ \\ &=1+2i+i^2&\small{\gray{\text{Simplify}}}\\ \\ &=1+2i+(-1)&\small{\gray{i^2=-1}}\\ \\ &=2i&\small{\gray{\text{Simplify}}} \end{aligned}

## Challenge Problems

### Problem 1

Let a and b be real numbers. What is left parenthesis, a, minus, b, i, right parenthesis, left parenthesis, a, plus, b, i, right parenthesis?

\begin{aligned}(a-bi) (a+bi)&=a(a)+a(bi)+({-bi})(a)+({-bi})(bi)&\small{\gray{\text{Multiply}}}\\ \\ &=a^2+abi-abi-b^2i^2&\small{\gray{\text{Simplify}}}\\ \\ &=a^2-b^2i^2&\small{\gray{\text{Simplify}}}\\ \\ &=a^2-b^2(-1)&\small{\gray{i^2=-1}}\\ \\ &=a^2+b^2&\small{\gray{\text{Simplify}}} \end{aligned}

### Problem 2

Perform the indicated operation and simplify. left parenthesis, 1, plus, 3, i, right parenthesis, start superscript, 2, end superscript, dot, left parenthesis, 2, plus, i, right parenthesis
\begin{aligned}(1+3i)^2\cdot (2+i)&=\greenD{(1+3i)(1+3i)}(2+i)\\ \\ &=\greenD{\left(1+3i+3i+9i^2\right)}(2+i)\\ \\ &=\greenD{\left(1+6i+9i^2\right)}(2+i)\\ \\ &=\greenD{\left(1+6i+9(-1)\right)}(2+i)\\ \\ &=\greenD{(-8+6i)}(2+i)\\ \\ &=-16-8i+12i+6i^2\\ \\ &=-16+4i+6(-1)\\ \\ &=-22+4i \end{aligned}