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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!
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 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.
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, 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.
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, plus, 6, i, and so we have that 2, i, left parenthesis, 3, minus, 8, i, right parenthesis, equals, 16, plus, 6, i.

Check your understanding

Problem 1

Multiply 3, left parenthesis, minus, 2, plus, 10, i, right parenthesis.
Write your answer in the form of a, plus, b, i.

Problem 2

Multiply minus, 6, i, left parenthesis, 5, plus, 7, i, right parenthesis.
Write your answer in the form of a, plus, b, i.

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.
(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 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.
(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}

Check your understanding

Problem 3

Multiply left parenthesis, 1, plus, 2, i, right parenthesis, left parenthesis, 3, plus, i, right parenthesis.
Write your answer in the form of a, plus, b, i.

Problem 4

Multiply left parenthesis, 4, plus, i, right parenthesis, left parenthesis, 7, minus, 3, i, right parenthesis.
Write your answer in the form of a, plus, b, i.

Problem 5

Multiply left parenthesis, 2, minus, i, right parenthesis, left parenthesis, 2, plus, i, right parenthesis.
Write your answer in the form of a, plus, b, i.

Problem 6

Multiply left parenthesis, 1, plus, i, right parenthesis, left parenthesis, 1, plus, i, right parenthesis.
Write your answer in the form of a, plus, b, i.

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
Write your answer in the form of a, plus, b, i.

Want to join the conversation?

  • ohnoes default style avatar for user Lawrence Yount
    how do you make things so much easier than school does?
    (50 votes)
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  • hopper cool style avatar for user 🤔 ᴄᴏᴅᴇᴅ ɢᴇɴɪᴜȿ 😎
    From what I understand, when added, multiplying, subtracting, and dividing, you think and act like the i is a variable. Is that correct?
    (25 votes)
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    • primosaur ultimate style avatar for user JeremiahJTReed
      "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.
      (17 votes)
  • duskpin tree style avatar for user eunyounguhm
    In challenge problem 2, isn't ( 1 + 3i )^2 equal to ( 1 + 9i^2 ) = ( 1 + 9*( -1 ) ) = ( 1 - 9 ) = -8?
    (5 votes)
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  • blobby green style avatar for user yucao
    Is it necessary to practice some divisions on complex numbers? If so does it follow the same general division rule?
    (6 votes)
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    • cacteye blue style avatar for user Jerry Nilsson
      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
      (18 votes)
  • purple pi teal style avatar for user Phạm Trần Minh Trí
    So every complex numbers form is a+bi ?
    (4 votes)
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  • duskpin sapling style avatar for user Chloe Cho
    Does any one know how to solve Challege Problems 1 and 2?? I'm a bit confused.
    (4 votes)
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    • leaf green style avatar for user kubleeka
      (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.
      (5 votes)
  • blobby green style avatar for user supermanlaebahlol
    How to solve (-2+I)(-2-I)
    (2 votes)
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    • starky ultimate style avatar for user Paul Miller
      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).
      (7 votes)
  • spunky sam orange style avatar for user Super learner 24
    For −6i(5+7i) what would you do after distributing
    and would you get -30i-42i^2
    (3 votes)
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  • blobby green style avatar for user nicholsty2
    how would i multiply a complex number with a variable. such as (2x-1)(x+4i)?
    (3 votes)
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  • blobby green style avatar for user Carol Scudero
    If your answer is 31+5i, why does it reject +31+5i?
    (2 votes)
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