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.

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

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

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.

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.

how would i multiply a complex number with a variable. such as (2x-1)(x+4i)?

You would apply the distributive property in the usual way, and end up with an expression involving both your variable and (possibly) i. In this case, you get (2x-1)(x+4i) =(2x-1)x+(2x-1)4i =2x²-x+8xi-4i

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).

Why is it like that when you can't add/subtract complex numbers from real numbers but can multiply it?

You _can_ add a real and complex number if addition is being defined for complex numbers (as reals are also complex). So, if I have to add 3 and 5 + 2i, I'd get 8 + 2i. But, if addition is defined for reals, then adding a real to a complex doesn't make sense, as well, that's not a defined operation.

Main content

Algebra (all content)

Course: Algebra (all content) > Unit 16

Lesson 5: Multiplying complex numbers

Multiplying complex numbers

CCSS.Math:

HSN.CN.A.2

HSN.CN.A

Google Classroom

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.

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.

[I need help!]

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.

[I need help!]

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}

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.

[I need help!]

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.

[I need help!]

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.

[I need help!]

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.

[I need help!]

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?

[I need help!]

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.

[I need help!]

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Lawrence Yount
Posted 3 years ago. Direct link to Lawrence Yount's post “how do you make things so...”
how do you make things so much easier than school does?
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(73 votes)
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- Rebecca
  Posted 3 years ago. Direct link to Rebecca's post “Sal explains the concepts...”
  Sal explains the concepts rather than just gets you to memorise a method.
  Comment on Rebecca's post “Sal explains the concepts...”
  (80 votes)
🤔 ᴄᴏᴅᴇᴅ ɢᴇɴɪᴜȿ 😎
Posted 6 years ago. Direct link to 🤔 ᴄᴏᴅᴇᴅ ɢᴇɴɪᴜȿ 😎's post “From what I understand, w...”
From what I understand, when added, multiplying, subtracting, and dividing, you think and act like the i is a variable. Is that correct?
Button navigates to signup pageComment on 🤔 ᴄᴏᴅᴇᴅ ɢᴇɴɪᴜȿ 😎's post “From what I understand, w...”
(27 votes)
Answer
- JeremiahJTReed
  Posted 5 years ago. Direct link to JeremiahJTReed's post “"i" will act as the varia...”
  "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.
  Button navigates to signup page
  (21 votes)
eunyounguhm
Posted 6 years ago. Direct link to eunyounguhm's post “In challenge problem 2, i...”
In challenge problem 2, isn't ( 1 + 3i )^2 equal to ( 1 + 9i^2 ) = ( 1 + 9*( -1 ) ) = ( 1 - 9 ) = -8?
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
- Kim Seidel
  Posted 6 years ago. Direct link to Kim Seidel's post “No... to multiply ( 1 + 3...”
  No... to multiply ( 1 + 3i )^2, you must use FOIL.
  It becomes 1 + 6i -9 = 6i - 8
  Comment on Kim Seidel's post “No... to multiply ( 1 + 3...”
  (28 votes)
yucao
Posted 6 years ago. Direct link to yucao's post “Is it necessary to practi...”
Is it necessary to practice some divisions on complex numbers? If so does it follow the same general division rule?
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(6 votes)
Answer
- Jerry Nilsson
  Posted 6 years ago. Direct link to Jerry Nilsson's post “When dividing two complex...”
  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
  Button navigates to signup page
  (21 votes)
Phạm Trần Minh Trí
Posted 3 years ago. Direct link to Phạm Trần Minh Trí's post “So every complex numbers ...”
So every complex numbers form is a+bi ?
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(4 votes)
Answer
- Ashish B
  Posted 2 years ago. Direct link to Ashish B's post “Yes... For a real number,...”
  Yes... For a real number, in a+bi, b=0
  For a imaginary number, in a+bi, a=0
  Button navigates to signup page
  (2 votes)
Chloe Cho
Posted 6 years ago. Direct link to Chloe Cho's post “Does any one know how to ...”
Does any one know how to solve Challege Problems 1 and 2?? I'm a bit confused.
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(4 votes)
Answer
- kubleeka
  Posted 6 years ago. Direct link to kubleeka's post “(a-bi)*(a+bi) We multiply...”
  (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.
  Comment on kubleeka's post “(a-bi)*(a+bi) We multiply...”
  (5 votes)
supermanlaebahlol
Posted 6 years ago. Direct link to supermanlaebahlol's post “How to solve (-2+I)(-2-I)”
How to solve (-2+I)(-2-I)
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(2 votes)
Answer
- Paul Miller
  Posted 6 years ago. Direct link to Paul Miller's post “Once you expand the binom...”
  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).
  Button navigates to signup page
  (7 votes)
Super learner 24
Posted 2 years ago. Direct link to Super learner 24's post “For −6i(5+7i) what would ...”
For −6i(5+7i) what would you do after distributing
and would you get -30i-42i^2
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(3 votes)
Answer
- Kim Seidel
  Posted 2 years ago. Direct link to Kim Seidel's post “So far, you're doing grea...”
  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.
  Comment on Kim Seidel's post “So far, you're doing grea...”
  (3 votes)
nicholsty2
Posted 3 years ago. Direct link to nicholsty2's post “how would i multiply a co...”
how would i multiply a complex number with a variable. such as (2x-1)(x+4i)?
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(3 votes)
Answer
- kubleeka
  Posted 3 years ago. Direct link to kubleeka's post “You would apply the distr...”
  You would apply the distributive property in the usual way, and end up with an expression involving both your variable and (possibly) i.
  
  In this case, you get (2x-1)(x+4i)
  =(2x-1)x+(2x-1)4i
  =2x²-x+8xi-4i
  Comment on kubleeka's post “You would apply the distr...”
  (2 votes)
Aveesh
Posted 2 months ago. Direct link to Aveesh's post “Why is it like that when ...”
Why is it like that when you can't add/subtract complex numbers from real numbers
but can multiply it?
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(2 votes)
Answer
- Venkata
  Posted 2 months ago. Direct link to Venkata's post “You _can_ add a real and ...”
  You can add a real and complex number if addition is being defined for complex numbers (as reals are also complex). So, if I have to add 3 and 5 + 2i, I'd get 8 + 2i.
  
  But, if addition is defined for reals, then adding a real to a complex doesn't make sense, as well, that's not a defined operation.
  Button navigates to signup page
  (3 votes)