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

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.

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 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!!

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

can I create my own complex numbers? like what if I created imaginary numbers called j

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.

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).

multiplying complex numbers is hard.

Multiplying complex number uses FOIL (which you already know how to do). Treat i as a variable. Then, you have a new step. One term will include i^2. Since i^2=-1, you need to swap out the i^2. For example: (2+3i)(5-2i) -- FOIL: 10 - 4i + 15i -6i^2 -- Turn i^2 into -1: 10 - 4i + 15i -6(-1) -- Simplify & combine like terms: 10 + 11i + 6 = 17+ 11i Hope this helps.

Main content

Course: Get ready for Precalculus > Unit 1

Lesson 5: Multiplying complex numbers

Multiplying complex numbers

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

a + b i

, 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 + 5 i)

. Write the resulting number in the form of

a + b i

Solution

If your instinct tells you to distribute the

- 4

, your instinct would be right! Let's do that!

\begin{aligned} - 4 (13 + 5 i) & = - 4 (13) + (- 4) (5 i) \\ = - 52 - 20 i \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 (3 - 8 i)

. Write the resulting number in the form of

a + b i

Solution

Again, let's start by distributing the

2 i

to each term in the parentheses.

\begin{aligned} 2 i (3 - 8 i) & = 2 i (3) - 2 i (8 i) \\ = 6 i - 16 i^{2} \end{aligned}

At this point, the answer is not of the form

a + b i

since it contains

i^{2}

However, we know that

i^{2} = - 1

. Let's substitute and see where that gets us.

\begin{aligned} = 6 i - 16 i^{2} \\ = 6 i - 16 (- 1) \\ = 6 i + 16 \end{aligned}

Using the commutative property, we can write the answer as

16 + 6 i

, and so we have that

2 i (3 - 8 i) = 16 + 6 i

Check your understanding

Problem 1

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

Problem 2

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

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

(1 + 4 i) (5 + i)

. Write the resulting number in the form of

a + 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} (1 + 4 i) (5 + i) & = (1) (5) + (1) (i) + (4 i) (5) + (4 i) (i) \\ = 5 + i + 20 i + 4 i^{2} \\ = 5 + 21 i + 4 i^{2} \end{aligned}

Since

i^{2} = - 1

, we can replace

i^{2}

with

- 1

to obtain the desired form of

a + b i

\begin{aligned} = 5 + 21 i + 4 i^{2} \\ = 5 + 21 i + 4 (- 1) \\ = 5 + 21 i - 4 \\ = 1 + 21 i \end{aligned}

Check your understanding

Problem 3

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

Problem 4

Multiply $(4 + i) (7 - 3 i)$ ‍.
Write your answer in the form of $a + b i$ ‍.

Problem 5

Multiply $(2 - i) (2 + i)$ ‍.
Write your answer in the form of $a + b i$ ‍.

Problem 6

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

Challenge Problems

Problem 1

Let $a$ ‍ and $b$ ‍ be real numbers. What is $(a - b i) (a + b i)$ ‍?

Problem 2

Perform the indicated operation and simplify. $(1 + 3 i)^{2} \cdot (2 + i)$ ‍
Write your answer in the form of $a + b i$ ‍.

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Lawrence Yount
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how do you make things so much easier than school does?
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- Rebecca
  Posted 4 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...”
  (144 votes)
🤔 ᴄᴏᴅᴇᴅ ɢᴇɴɪᴜȿ 😎
Posted 7 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?
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(36 votes)
Answer
- JeremiahJTReed
  Posted 6 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
  (30 votes)
yucao
Posted 7 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?
Button navigates to signup pageButton navigates to signup page
(7 votes)
Answer
- Jerry Nilsson
  Posted 7 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
  (27 votes)
eunyounguhm
Posted 7 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
(3 votes)
Answer
- Kim Seidel
  Posted 7 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...”
  (30 votes)
joshuaterefe73
Posted 9 months ago. Direct link to joshuaterefe73's post “can I create my own compl...”
can I create my own complex numbers? like what if I created imaginary numbers called j
Button navigates to signup pageComment on joshuaterefe73's post “can I create my own compl...”
(2 votes)
Answer
- GemMarkle
  Posted 8 months ago. Direct link to GemMarkle's post “Good mathematical communi...”
  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.
  Button navigates to signup page
  (3 votes)
Phạm Trần Minh Trí
Posted 4 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 3 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
  (3 votes)
Chloe Cho
Posted 7 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.
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- kubleeka
  Posted 7 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...”
  (6 votes)
supermanlaebahlol
Posted 7 years ago. Direct link to supermanlaebahlol's post “How to solve (-2+I)(-2-I)”
How to solve (-2+I)(-2-I)
Button navigates to signup pageComment on supermanlaebahlol's post “How to solve (-2+I)(-2-I)”
(2 votes)
Answer
- Paul Miller
  Posted 7 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
  (9 votes)
Vibhaan Reddy
Posted 3 months ago. Direct link to Vibhaan Reddy's post “So if you add a + bi you ...”
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!!
Button navigates to signup pageComment on Vibhaan Reddy's post “So if you add a + bi you ...”
(3 votes)
Answer
- FreeRadical
  Posted 3 months ago. Direct link to FreeRadical's post “Yes that is correct! the ...”
  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
  Comment on FreeRadical's post “Yes that is correct! the ...”
  (2 votes)
rehadam31
Posted a year ago. Direct link to rehadam31's post “multiplying complex numbe...”
multiplying complex numbers is hard.
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Kim Seidel
  Posted a year ago. Direct link to Kim Seidel's post “Multiplying complex numbe...”
  Multiplying complex number uses FOIL (which you already know how to do). Treat i as a variable. Then, you have a new step. One term will include i^2. Since i^2=-1, you need to swap out the i^2. For example:
  (2+3i)(5-2i)
  -- FOIL: 10 - 4i + 15i -6i^2
  -- Turn i^2 into -1: 10 - 4i + 15i -6(-1)
  -- Simplify & combine like terms: 10 + 11i + 6 = 17+ 11i
  
  Hope this helps.
  Comment on Kim Seidel's post “Multiplying complex numbe...”
  (4 votes)