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.

parents check your grade and send you to khan academy😰

strange, i created this account myself cause i wanted to learn meths

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

get harder but I am fine

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

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

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?

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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...”
(27 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://...”
  (28 votes)
dylan.forr99
Posted 3 months ago. Direct link to dylan.forr99's post “Honestly, when I arrived ...”
Honestly, when I arrived at this unit, I was very intimidated, but it's really not as bad as I imagined so far. Thanks Sal!
Button navigates to signup pageComment on dylan.forr99's post “Honestly, when I arrived ...”
(28 votes)
Answer
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 8 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...”
  (5 votes)
nikhilsivajiss
Posted 3 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...”
(7 votes)
Answer
- khushpa5
  Posted 3 months ago. Direct link to khushpa5'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
  (10 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...”
  (9 votes)
The ŇØŦ€ŞŁΔ¥€Ř
Posted 4 months ago. Direct link to The ŇØŦ€ŞŁΔ¥€Ř's post “parents check your grade ...”
parents check your grade and send you to khan academy😰
Button navigates to signup pageComment on The ŇØŦ€ŞŁΔ¥€Ř's post “parents check your grade ...”
(6 votes)
Answer
- kitty-chan
  Posted 2 months ago. Direct link to kitty-chan's post “strange, i created this a...”
  strange, i created this account myself cause i wanted to learn meths
  Button navigates to signup page
  (3 votes)
Abisade Adegboye
Posted 7 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
(2 votes)
Answer
- Kim Seidel
  Posted 7 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
  (6 votes)
Michael Batizaze
Posted 2 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
(4 votes)
Answer
- Kevin
  Posted 2 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 ...”
  (2 votes)
Lumiknoll
Posted a month ago. Direct link to Lumiknoll's post “super duper very cool”
super duper very cool
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(4 votes)
Answer