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

Solution

If your instinct tells you to distribute the 4-4, your instinct would be right! Let's do that!
4(13+5i)=4(13)+(4)(5i)=5220i\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(38i)2i (3-8i). Write the resulting number in the form of a+bia+bi.

Solution

Again, let's start by distributing the 2i2i to each term in the parentheses.
2i(38i)=2i(3)2i(8i)=6i16i2\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+bia+bi since it contains i2i^2.
However, we know that i2=1\goldD{i^2=-1}. Let's substitute and see where that gets us.
2i(38i)=6i16i2=6i16(1)=6i+16\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+6i16+6i, and so we have that 2i(38i)=16+6i2i (3-8i)=16+6i.

Check your understanding

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

Solution

In this example, some find it very helpful to think of ii 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.
(1+4i)(5+i)=(1)(5)+(1)(i)+(4i)(5)+(4i)(i)=5+i+20i+4i2=5+21i+4i2\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 i2=1\goldD{i^2=-1}, we can replace i2i^2 with 1-1 to obtain the desired form of a+bia+bi.
(15i)(6+i)=5+21i+4i2=5+21i+4(1)=5+21i4=1+21i\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}

Tjek din forståelse

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Problem 5

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Challenge Problems

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