What about dividing complex numbers?

Go to this link: https://www.khanacademy.org/math/precalculus/imaginary-and-complex-numbers/complex-conjugates-and-dividing-complex-numbers/v/dividing-complex-numbers

When is a practical situation to apply imaginary numbers?

rotations like how you have i, -1, -i, and +1. Look into radio signaling or transmissions, i think it applies to quantum as well.

How do you get Multiplication on the table above?

Use Distributive Property or FOIL to multiply two binomials together. Combine like terms. Factor out i from the imaginary terms. Remember that i squared equals -1.

Evaluate 1+i+i^2+i^3+...+I,^1000

This is kinda easy. We know that the patter of i^n is: ``` n: 0 1 2 3 4 5 6 7... i^n: 1 i -1 -i 1 i -1 -i...``` we can see from the pattern above,[1, i, -1, -i] cycles on and on. There are 1000/4 = 250 cycles of [1, i, -1, -i] in this question. The sum of the cycles (1, i, -1, -i) is 0. Therefore, the answere is 250*0 = 0. ANSWERE: 1+i+i^2+i^3+...+i^1000 = 0.

get harder but I am fine

It's going to get harder the father down the road you go.

What do you do when a number is put to the power of i? For example 2^i

You will need Euler's formula, e^(ix)=cos(x)+i•sin(x), which is derived in the calculus playlist. This identity is true for every real or complex x. If you take this as given, we can derive 2^i=e^(ln(2^i))=e^(i•ln(2)) with logarithm properties. Then by Euler's formula, e^(i•ln(2))=cos(ln(2))+i•sin(ln(2))≈0.769+0.639i

I think there's an error with the Multiplication formula (if I'd call it that) It should be (a1a2+b1b2)... not (a1a2-b1b2)... correct me if I'm wrong though

Sorry, the formula is correct. (b1)i*(b2i) = b1b2i^2 Since i^2 = -1, the b1b2 switches negative. Hope this helps.

I'm still a little confused about multiplying complex numbers like (−3−2i)⋅(−4+2i). I got the jist of subtracting and adding, but I keep getting it wrong for multiplying. HeLp mE, pLeAsE.

Well, the thing with multiplying complex numbers is what you do with the "i". In this particular example you can use FOIL method here: (-3*-4)+(-3*2i)+(-2i*-4)+(-2i*2i) Now, this simplifies to 12-6i+8i+4. Do you see why the last term is positive 4 and not negative 4? i*i= -1. So the answer would further simplify to 16+2i.

Can you please help with this question? :( (2 + i)(3 - i)(1 + 2i)(1 - i)(3 + i)

You will have to use the FOIL method. Treat i as you would a variable, and pick two parantheses. Multiply the first terms, then the outside, then the inside, and then the last terms. Once you do that, multiply the rest of the factors out until there is nothing left to multiply out. Combine like terms. Before you are finished, remember that i² = -1, so substitute -1 everywhere you see an i² and simplify further. This should give you a final and simplified answer.

Main content

Course: Algebra 2 > Unit 2

Lesson 5: Multiplying complex numbers

Complex number operations review

Google Classroom

Review complex number addition, subtraction, and multiplication.


Addition
$(a_{1} + b_{1} i) + (a_{2} + b_{2} i) = (a_{1} + a_{2}) + (b_{1} + b_{2}) i$ ‍
Subtraction
$(a_{1} + b_{1} i) - (a_{2} + b_{2} i) = (a_{1} - a_{2}) + (b_{1} - b_{2}) i$ ‍
Multiplication
$(a_{1} + b_{1} i) \cdot (a_{2} + b_{2} i) = (a_{1} a_{2} - b_{1} b_{2}) + (a_{1} b_{2} + a_{2} b_{1}) i$ ‍

Want to learn more about complex number operations? Check out these videos:

Adding complex numbers
Subtracting complex numbers
Multiplying complex numbers

Practice set 1: Adding and subtracting complex numbers

Example 1: Adding complex numbers

When adding complex numbers, we simply add the real parts and add the imaginary parts. For example:

\begin{aligned} (3 + 4 i) + (6 - 10 i) \\ = (3 + 6) + (4 - 10) i \\ = 9 - 6 i \end{aligned}

Example 2: Subtracting complex numbers

When subtracting complex numbers, we simply subtract the real parts and subtract the imaginary parts. For example:

\begin{aligned} (3 + 4 i) - (6 - 10 i) \\ = (3 - 6) + (4 - (- 10)) i \\ = - 3 + 14 i \end{aligned}

Problem 1.1

(7 - 10 i) - (3 + 30 i) =

Express your answer in the form $(a + b i)$ ‍.

Want to try more problems like this? Check out this exercise.

Practice set 2: Multiplying complex numbers

When multiplying complex numbers, we perform a multiplication similar to how we expand the parentheses in binomial products:

(a + b) (c + d) = a c + a d + b c + b d

Unlike regular binomial multiplication, with complex numbers we also consider the fact that

i^{2} = - 1

Example 1

\begin{aligned} 2 \cdot (- 3 + 4 i) \\ = 2 \cdot (- 3) + 2 \cdot 4 i \\ = - 6 + 8 i \end{aligned}

Example 2

\begin{aligned} 3 i \cdot (1 - 5 i) \\ = 3 i \cdot 1 + 3 i \cdot (- 5) i \\ = 3 i - 15 i^{2} \\ = 3 i - 15 (- 1) \\ = 15 + 3 i \end{aligned}

Example 3

\begin{aligned} (2 + 3 i) \cdot (1 - 5 i) \\ = 2 \cdot 1 + 2 \cdot (- 5) i + 3 i \cdot 1 + 3 i \cdot (- 5) i \\ = 2 - 10 i + 3 i - 15 i^{2} \\ = 2 - 7 i - 15 (- 1) \\ = 17 - 7 i \end{aligned}

Problem 2.1

8 \cdot (11 i + 2) =

Your answer should be a complex number in the form $a + b i$ ‍ where $a$ ‍ and $b$ ‍ are real numbers.

Want to try more problems like this? Check out this basic exercise and this advanced exercise.

Want to join the conversation?

Sort by:

Wanli Tan
Posted 7 years ago. Direct link to Wanli Tan's post “What about dividing compl...”
What about dividing complex numbers?
Button navigates to signup pageComment on Wanli Tan's post “What about dividing compl...”
(40 votes)
Answer
- ThelegendarySupersayiongod
  Posted 7 years ago. Direct link to ThelegendarySupersayiongod's post “Go to this link: https://...”
  Go to this link: https://www.khanacademy.org/math/precalculus/imaginary-and-complex-numbers/complex-conjugates-and-dividing-complex-numbers/v/dividing-complex-numbers
  Comment on ThelegendarySupersayiongod's post “Go to this link: https://...”
  (45 votes)
nissru
Posted 7 years ago. Direct link to nissru's post “Evaluate 1+i+i^2+i^3+...+...”
Evaluate 1+i+i^2+i^3+...+I,^1000
Button navigates to signup pageComment on nissru's post “Evaluate 1+i+i^2+i^3+...+...”
(7 votes)
Answer
- He Ivan
  Posted a year ago. Direct link to He Ivan's post “This is kinda easy. We kn...”
  This is kinda easy.
  We know that the patter of i^n is:
  n: 0 1 2 3 4 5 6 7... i^n: 1 i -1 -i 1 i -1 -i...
  
  we can see from the pattern above,[1, i, -1, -i] cycles on and on.
  There are 1000/4 = 250 cycles of [1, i, -1, -i] in this question.
  The sum of the cycles (1, i, -1, -i) is 0.
  Therefore, the answere is 250*0 = 0.
  ANSWERE: 1+i+i^2+i^3+...+i^1000 = 0.
  Comment on He Ivan's post “This is kinda easy. We kn...”
  (9 votes)
nikhilsivajiss
Posted 10 months ago. Direct link to nikhilsivajiss's post “When is a practical situa...”
When is a practical situation to apply imaginary numbers?
Button navigates to signup pageComment on nikhilsivajiss's post “When is a practical situa...”
(9 votes)
Answer
- Khush P.
  Posted 10 months ago. Direct link to Khush P.'s post “rotations like how you ha...”
  rotations like how you have i, -1, -i, and +1. Look into radio signaling or transmissions, i think it applies to quantum as well.
  Button navigates to signup page
  (13 votes)
Yura Er
Posted 8 years ago. Direct link to Yura Er's post “How do you get Multiplica...”
How do you get Multiplication on the table above?
Button navigates to signup pageComment on Yura Er's post “How do you get Multiplica...”
(8 votes)
Answer
- Belinda Cheng
  Posted 7 years ago. Direct link to Belinda Cheng's post “Use Distributive Property...”
  Use Distributive Property or FOIL to multiply two binomials together.
  Combine like terms.
  Factor out i from the imaginary terms.
  Remember that i squared equals -1.
  Button navigates to signup page
  (9 votes)
Jimena Garcia
Posted 6 years ago. Direct link to Jimena Garcia's post “What do you do when a num...”
What do you do when a number is put to the power of i?
For example 2^i
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
- kubleeka
  Posted 6 years ago. Direct link to kubleeka's post “You will need Euler's for...”
  You will need Euler's formula, e^(ix)=cos(x)+i•sin(x), which is derived in the calculus playlist. This identity is true for every real or complex x.
  
  If you take this as given, we can derive 2^i=e^(ln(2^i))=e^(i•ln(2)) with logarithm properties.
  
  Then by Euler's formula, e^(i•ln(2))=cos(ln(2))+i•sin(ln(2))≈0.769+0.639i
  Comment on kubleeka's post “You will need Euler's for...”
  (10 votes)
Lumiknoll
Posted 8 months ago. Direct link to Lumiknoll's post “super duper very cool”
super duper very cool
Button navigates to signup pageButton navigates to signup page
(9 votes)
Answer
Michael Batizaze
Posted 9 months ago. Direct link to Michael Batizaze's post “get harder but I am fine”
get harder but I am fine
Button navigates to signup pageButton navigates to signup page
(7 votes)
Answer
- Kevin
  Posted 8 months ago. Direct link to Kevin's post “It's going to get harder ...”
  It's going to get harder the father down the road you go.
  Comment on Kevin's post “It's going to get harder ...”
  (5 votes)
Abisade Adegboye
Posted a year ago. Direct link to Abisade Adegboye's post “I think there's an error ...”
I think there's an error with the Multiplication formula (if I'd call it that)
It should be (a1a2+b1b2)...
not (a1a2-b1b2)...
correct me if I'm wrong though
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Kim Seidel
  Posted a year ago. Direct link to Kim Seidel's post “Sorry, the formula is cor...”
  Sorry, the formula is correct. (b1)i*(b2i) = b1b2i^2
  Since i^2 = -1, the b1b2 switches negative.
  Hope this helps.
  Button navigates to signup page
  (9 votes)
Anishka Rahul
Posted 5 years ago. Direct link to Anishka Rahul's post “I'm still a little confus...”
I'm still a little confused about multiplying complex numbers like (−3−2i)⋅(−4+2i). I got the jist of subtracting and adding, but I keep getting it wrong for multiplying. HeLp mE, pLeAsE.
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Ron Joniak
  Posted 5 years ago. Direct link to Ron Joniak's post “Well, the thing with mult...”
  Well, the thing with multiplying complex numbers is what you do with the "i". In this particular example you can use FOIL method here:
  (-3*-4)+(-3*2i)+(-2i*-4)+(-2i*2i)
  Now, this simplifies to 12-6i+8i+4. Do you see why the last term is positive 4 and not negative 4? i*i= -1.
  So the answer would further simplify to 16+2i.
  Comment on Ron Joniak's post “Well, the thing with mult...”
  (5 votes)
Evin Nazya
Posted 6 years ago. Direct link to Evin Nazya's post “Can you please help with ...”
Can you please help with this question? :(
(2 + i)(3 - i)(1 + 2i)(1 - i)(3 + i)
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Ian
  Posted 6 years ago. Direct link to Ian's post “You will have to use the ...”
  You will have to use the FOIL method. Treat i as you would a variable, and pick two parantheses. Multiply the first terms, then the outside, then the inside, and then the last terms. Once you do that, multiply the rest of the factors out until there is nothing left to multiply out. Combine like terms. Before you are finished, remember that i² = -1, so substitute -1 everywhere you see an i² and simplify further. This should give you a final and simplified answer.
  Comment on Ian's post “You will have to use the ...”
  (4 votes)