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

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.

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'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.

can someone explain how this equation is true (a1+b1i)⋅(a2+b2i)= (a1a2−b1b2)+(a1b2+a2b1)i

You have 2 binomials to multiply... Use FOIL or extended distribution. Every term in the first binomial is multiplied with each term in the 2nd binomial. (a1+b1i)⋅(a2+b2i) = a1a2+a1b2i+a2b1i+b1b2i^2 Simplify the i^2 into -1 and the b1b2i^2 becomes =b1b2 = a1a2+a1b2i+a2b1i-b1b2 Use commutative property to reorder the terms = a1a2-b1b2+a1b2i+a2b1i Use associative property to add parentheses = (a1a2-b1b2)+(a1b2i+a2b1i) Factor out i from the last 2 terms = (a1a2-b1b2)+(a1b2+a2b1)i Hope this helps.

What about 6 ( -7 + 6i ) ( -4 + 2i ) ?

Use FOIL or extended distribution to multiply: ( -7 + 6i ) ( -4 + 2i ) Then, distribute 6 across the result. Hope this helps.

Hi! Great lesson. Where would I be able to find videos simplifying large powers of i? For example, i^89... Thank you!

Use the search bar at the top of any KhanAcademy screen and search for "powers of i". There is a video that explains how to deal with problems like you've described.

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.

Main content

Algebra 2

Course: Algebra 2 > Unit 2

Lesson 5: Multiplying complex numbers

Complex number operations review

CCSS.Math:

HSN.CN.A.2

HSN.CN.A

Google Classroom

Review complex number addition, subtraction, and multiplication.


Addition
left parenthesis, a, start subscript, 1, end subscript, plus, b, start subscript, 1, end subscript, i, right parenthesis, plus, left parenthesis, a, start subscript, 2, end subscript, plus, b, start subscript, 2, end subscript, i, right parenthesis, equals, left parenthesis, a, start subscript, 1, end subscript, plus, a, start subscript, 2, end subscript, right parenthesis, plus, left parenthesis, b, start subscript, 1, end subscript, plus, b, start subscript, 2, end subscript, right parenthesis, i
Subtraction
left parenthesis, a, start subscript, 1, end subscript, plus, b, start subscript, 1, end subscript, i, right parenthesis, minus, left parenthesis, a, start subscript, 2, end subscript, plus, b, start subscript, 2, end subscript, i, right parenthesis, equals, left parenthesis, a, start subscript, 1, end subscript, minus, a, start subscript, 2, end subscript, right parenthesis, plus, left parenthesis, b, start subscript, 1, end subscript, minus, b, start subscript, 2, end subscript, right parenthesis, i
Multiplication
left parenthesis, a, start subscript, 1, end subscript, plus, b, start subscript, 1, end subscript, i, right parenthesis, dot, left parenthesis, a, start subscript, 2, end subscript, plus, b, start subscript, 2, end subscript, i, right parenthesis, equals, left parenthesis, a, start subscript, 1, end subscript, a, start subscript, 2, end subscript, minus, b, start subscript, 1, end subscript, b, start subscript, 2, end subscript, right parenthesis, plus, left parenthesis, a, start subscript, 1, end subscript, b, start subscript, 2, end subscript, plus, a, start subscript, 2, end subscript, b, start subscript, 1, end subscript, right parenthesis, 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} &\phantom{=}(\blueD 3+\greenD4i)+(\blueD6\greenD{-10}i) \\\\ &=(\blueD3+\blueD6)+(\greenD4\greenD{-10})i \\\\ &=\blueD9\greenD{-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} &\phantom{=}(\blueD 3+\greenD4i)-(\blueD6\greenD{-10}i) \\\\ &=(\blueD3-\blueD6)+(\greenD4-(\greenD{-10}))i \\\\ &=\blueD{-3}+\greenD{14}i \end{aligned}

Problem 1.1

Current

left parenthesis, 7, minus, 10, i, right parenthesis, minus, left parenthesis, 3, plus, 30, i, right parenthesis, equals

Express your answer in the form left parenthesis, a, plus, b, i, right parenthesis.

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:

left parenthesis, a, plus, b, right parenthesis, left parenthesis, c, plus, d, right parenthesis, equals, a, c, plus, a, d, plus, b, c, plus, b, d

Unlike regular binomial multiplication, with complex numbers we also consider the fact that i, squared, equals, minus, 1.

Example 1

\begin{aligned} &\phantom{=}\blueD 2\cdot(\blueD{-3}+\greenD{4}i) \\\\ &=\blueD2\cdot(\blueD{-3})+\blueD2\cdot\greenD4i \\\\ &=\blueD{-6}+\greenD8i \end{aligned}

Example 2

\begin{aligned} &\phantom{=}\greenD3i\cdot(\blueD{1}\greenD{-5}i) \\\\ &=\greenD3i\cdot\blueD1+\greenD3i\cdot(\greenD{-5})i \\\\ &=\greenD3i-15i^2 \\\\ &=\greenD3i-15(-1) \\\\ &=\blueD{15}+\greenD3i \end{aligned}

Example 3

\begin{aligned} &\phantom{=}(\blueD2+\greenD3i)\cdot(\blueD{1}\greenD{-5}i) \\\\ &=\blueD2\cdot\blueD1+\blueD2\cdot(\greenD{-5})i+\greenD3i\cdot\blueD1+\greenD3i\cdot(\greenD{-5})i \\\\ &=\blueD2\greenD{-10}i+\greenD3i-15i^2 \\\\ &=\blueD2\greenD{-7}i-15(-1) \\\\ &=\blueD{17}\greenD{-7}i \end{aligned}

Problem 2.1

Current

8, dot, left parenthesis, 11, i, plus, 2, right parenthesis, equals

Your answer should be a complex number in the form a, plus, 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 6 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...”
(26 votes)
Answer
- ThelegendarySupersayiongod
  Posted 6 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://...”
  (25 votes)
nissru
Posted 6 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+...+...”
(4 votes)
Answer
- He Ivan
  Posted 4 months 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...”
  (3 votes)
Yura Er
Posted 7 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...”
(6 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
  (7 votes)
Jimena Garcia
Posted 5 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
(4 votes)
Answer
- kubleeka
  Posted 5 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...”
  (7 votes)
Anishka Rahul
Posted 4 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
(2 votes)
Answer
- Ron Joniak
  Posted 4 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...”
  (4 votes)
Evin Nazya
Posted 5 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 Foreman
  Posted 5 years ago. Direct link to Ian Foreman'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 Foreman's post “You will have to use the ...”
  (3 votes)
Islam Hamdi
Posted 4 years ago. Direct link to Islam Hamdi's post “can someone explain how t...”
can someone explain how this equation is true
(a1+b1i)⋅(a2+b2i)= (a1a2−b1b2)+(a1b2+a2b1)i
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Kim Seidel
  Posted 4 years ago. Direct link to Kim Seidel's post “You have 2 binomials to m...”
  You have 2 binomials to multiply... Use FOIL or extended distribution. Every term in the first binomial is multiplied with each term in the 2nd binomial.
  (a1+b1i)⋅(a2+b2i) = a1a2+a1b2i+a2b1i+b1b2i^2
  Simplify the i^2 into -1 and the b1b2i^2 becomes =b1b2
  = a1a2+a1b2i+a2b1i-b1b2
  Use commutative property to reorder the terms
  = a1a2-b1b2+a1b2i+a2b1i
  Use associative property to add parentheses
  = (a1a2-b1b2)+(a1b2i+a2b1i)
  Factor out i from the last 2 terms
  = (a1a2-b1b2)+(a1b2+a2b1)i
  
  Hope this helps.
  Comment on Kim Seidel's post “You have 2 binomials to m...”
  (5 votes)
jasontkd2001
Posted 6 years ago. Direct link to jasontkd2001's post “What about 6 ( -7 + 6i ) ...”
What about 6 ( -7 + 6i ) ( -4 + 2i ) ?
Button navigates to signup pageComment on jasontkd2001's post “What about 6 ( -7 + 6i ) ...”
(1 vote)
Answer
- Kim Seidel
  Posted 6 years ago. Direct link to Kim Seidel's post “Use FOIL or extended dist...”
  Use FOIL or extended distribution to multiply: ( -7 + 6i ) ( -4 + 2i )
  Then, distribute 6 across the result.
  Hope this helps.
  Button navigates to signup page
  (5 votes)
Charlyn Otto
Posted 8 months ago. Direct link to Charlyn Otto's post “Hi! Great lesson. Where w...”
Hi! Great lesson. Where would I be able to find videos simplifying large powers of i? For example, i^89... Thank you!
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Kim Seidel
  Posted 8 months ago. Direct link to Kim Seidel's post “Use the search bar at the...”
  Use the search bar at the top of any KhanAcademy screen and search for "powers of i". There is a video that explains how to deal with problems like you've described.
  Button navigates to signup page
  (2 votes)
Abisade Adegboye
Posted 3 months 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
(1 vote)
Answer
- Kim Seidel
  Posted 3 months 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
  (2 votes)