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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 a+bi, where i is the imaginary unit and a and 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!
4(13+5i)=4(13)+(4)(5i)=5220i
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). 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.
2i(38i)=2i(3)2i(8i)=6i16i2
At this point, the answer is not of the form a+bi since it contains i2.
However, we know that i2=1. Let's substitute and see where that gets us.
2i(38i)=6i16i2=6i16(1)=6i+16
Using the commutative property, we can write the answer as 16+6i, and so we have that 2i(38i)=16+6i.

Check your understanding

Problem 1

Multiply 3(2+10i).
Write your answer in the form of a+bi.

Problem 2

Multiply 6i(5+7i).
Write your answer in the form of a+bi.

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.
(1+4i)(5+i)=(1)(5)+(1)(i)+(4i)(5)+(4i)(i)=5+i+20i+4i2=5+21i+4i2
Since i2=1, we can replace i2 with 1 to obtain the desired form of a+bi.
(15i)(6+i)=5+21i+4i2=5+21i+4(1)=5+21i4=1+21i

Check your understanding

Problem 3

Multiply (1+2i)(3+i).
Write your answer in the form of a+bi.

Problem 4

Multiply (4+i)(73i).
Write your answer in the form of a+bi.

Problem 5

Multiply (2i)(2+i).
Write your answer in the form of a+bi.

Problem 6

Multiply (1+i)(1+i).
Write your answer in the form of a+bi.

Challenge Problems

Problem 1

Let a and b be real numbers. What is (abi)(a+bi)?

Problem 2

Perform the indicated operation and simplify. (1+3i)2(2+i)
Write your answer in the form of a+bi.

Want to join the conversation?

  • ohnoes default style avatar for user Lawrence Yount
    how do you make things so much easier than school does?
    (111 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?
    (36 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?
    (7 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
      (25 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?
    (3 votes)
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  • blobby green style avatar for user joshuaterefe73
    can I create my own complex numbers? like what if I created imaginary numbers called j
    (3 votes)
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    • piceratops ultimate style avatar for user GemMarkle
      Good mathematical communication allows you to do any number of things. The convention is for imaginary numbers always to be represented by 'i'. However, if you began describing a mathematical model by saying, "Let 'j' equal the square root of -1 ..." everybody would understand what you meant. It can be just a variable, so long as you've defined it well, and it would continue to follow all the same rules of complex numbers.
      (4 votes)
  • 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.
    (5 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.
      (7 votes)
  • blobby green style avatar for user supermanlaebahlol
    How to solve (-2+I)(-2-I)
    (3 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).
      (9 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 tree style avatar for user Anthony Zazueta
    In what universe am I going to apply ANY OF THIS to an actual job? Because I doubt a Wal-Mart or Cargo Driver would need to actually apply this.
    (1 vote)
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    • piceratops sapling style avatar for user (YC) DoD
      Well, it obviously all depends on what job you find desirable... if you want to become a Walmart employee or a Cargo driver then you most certainly will not utilize complex numbers in your job.

      However, if you want to pursue a higher education and get a really nice job, then you have to learn about complex numbers. By learning about complex numbers today, you can make sure that you do well on your SAT/ACT or whatever college placement test you have in your country. Also, many jobs such as mathematician, physicist, quantum scientist, and computer scientists, all use complex and imaginary numbers. In fact, some of the most fundamental properties of the universe can only be defined with the imaginary number i... how could you get any more practical than that?

      Hope this helps motivate you :D
      (7 votes)
  • blobby green style avatar for user Vibhaan Reddy
    So if you add a + bi you can't alter anything, but if you multiply a * bi it becomes abi?? I just want clarification, thank you!!
    (3 votes)
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    • starky sapling style avatar for user FreeRadical
      Yes that is correct! the complex number i follows the same rules we learned in algebra when working with variables (like x). Only if the two terms belong to the same "family" can we add them.
      Example:
      2x + 3x + 4 = 5x + 4
      i + 5i + 2 = 6i + 2
      But when we multiply the x term by a number we can do the operation.
      Example:
      2x * 3 = 6x
      i*3 = 3i
      4*3i = 12i

      I hope this helps!
      FreeRadical
      (2 votes)