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## College Algebra

### Course: College Algebra>Unit 6

Lesson 3: Multiplying complex numbers

# 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 #1fab54, a, end color #1fab54, plus, start color #11accd, b, end color #11accd, i, where i is the imaginary unit and start color #1fab54, a, end color #1fab54 and start color #11accd, b, end color #11accd 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, squared.
However, we know that start color #e07d10, i, squared, equals, minus, 1, end color #e07d10. 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.

### Problem 2

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

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 #e07d10, i, squared, equals, minus, 1, end color #e07d10, we can replace i, squared 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.

### Problem 4

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

### Problem 5

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

### Problem 6

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

## 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?

### Problem 2

Perform the indicated operation and simplify. left parenthesis, 1, plus, 3, i, right parenthesis, squared, dot, left parenthesis, 2, plus, i, right parenthesis

## Want to join the conversation?

• how do you make things so much easier than school does?
• Sal explains the concepts rather than just gets you to memorise a method.
• From what I understand, when added, multiplying, subtracting, and dividing, you think and act like the i is a variable. Is that correct?
• "i" will act as the variable while you simplify the problem. But then you have to make sure to solve for "i" when possible. Like when you get i^2 or i^9 - make sure you solve for that instead of just leaving it as-is like you would a normal variable.
• In challenge problem 2, isn't ( 1 + 3i )^2 equal to ( 1 + 9i^2 ) = ( 1 + 9*( -1 ) ) = ( 1 - 9 ) = -8?
• No... to multiply ( 1 + 3i )^2, you must use FOIL.
It becomes 1 + 6i -9 = 6i - 8
• Is it necessary to practice some divisions on complex numbers? If so does it follow the same general division rule?
• When dividing two complex numbers in rectangular form we multiply the numerator and denominator by the complex conjugate of the denominator, because this effectively turns the denominator into a real number and the numerator becomes a multiplication of two complex numbers, which we can simplify.
The complex conjugate of (𝑎 + 𝑏𝑖) is (𝑎 − 𝑏𝑖).

Example:
(2 − 16𝑖) ∕ (5 − 𝑖) =
= (2 − 16𝑖) ∙ (5 + 𝑖) ∕ ((5 − 𝑖) ∙ (5 + 𝑖)) =
= (10 + 2𝑖 − 80𝑖 + 16) ∕ (25 + 5𝑖 − 5𝑖 + 1) =
= (26 − 78𝑖) ∕ 26 =
= 1 − 3𝑖

– – –

Alternatively, we can let
(2 − 16𝑖) = (5 − 𝑖)(𝑎 + 𝑏𝑖) = 5𝑎 + 5𝑏𝑖 −𝑎𝑖 + 𝑏 = (5𝑎 + 𝑏) + (5𝑏 − 𝑎)𝑖
and then solve the system of equations
5𝑎 + 𝑏 = 2
5𝑏 − 𝑎 = −16

𝑏 = 2 − 5𝑎
5(2 − 5𝑎) − 𝑎 = 10 − 26𝑎 = −16

𝑎 = 1
𝑏 = −3
• So every complex numbers form is a+bi ?
• Yes... For a real number, in a+bi, b=0
For a imaginary number, in a+bi, a=0
• Does any one know how to solve Challege Problems 1 and 2?? I'm a bit confused.
• (a-bi)*(a+bi)
We multiply this like any other binomial: we apply the distributive property twice to get
(a-bi)*a +(a-bi)*bi
a^2-abi +abi -(b^2)*(i^2)
The middle terms cancel and we get a^2 -(b^2)*(i^2)
Remember i^2=-1 and we get a^2 -(b^2)*(-1)
a^2+b^2

For the second one, try expanding the squared term first, simplifying it, then multiplying the second term.
• How to solve (-2+I)(-2-I)
• Once you expand the binomial, you will have two real terms and two imaginary terms (the i squared term is a real term since i^2=-1). THen you combine like terms. Since the two numbers you wrote are "conjugates" of each other, the imaginary terms will be opposites of each other and your answer will just be the real number (-2)^2 + 1^2 = 4 + 1 =5. You need to begin to recognize when you are multiplying conjugates as they result in a difference of squares (which for complex conjugates results in a sum of squares).
• For −6i(5+7i) what would you do after distributing
and would you get -30i-42i^2
• So far, you're doing great. You have one more thing to do.
i^2 = -1
So, use that to simplify the last term
-30i - 42(-1) = -30i + 42

Hope this helps.