# Multiplikation af komplekse tal

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 $\greenD{a}+\blueD{b}i$, where $i$ is the imaginary unit and $\greenD{a}$ and $\blueD{b}$ 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 $-4 (13+5i)$. Write the resulting number in the form of $a+bi$.

### Solution

If your instinct tells you to distribute the $-4$, 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 $2i (3-8i)$. Write the resulting number in the form of $a+bi$.

### Solution

Again, let's start by distributing the $2i$ 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 $a+bi$ since it contains $i^2$.
However, we know that $\goldD{i^2=-1}$. 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 $16+6i$, and so we have that $2i (3-8i)=16+6i$.

### Opgave 2

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 $(1+4i) (5+i)$. Write the resulting number in the form of $a+bi$.

### 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 $\goldD{i^2=-1}$, we can replace $i^2$ with $-1$ to obtain the desired form of $a+bi$.
\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}