If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Algebra 2

### Course: Algebra 2>Unit 2

Lesson 5: Multiplying complex numbers

# Complex number operations review

Review complex number addition, subtraction, and multiplication.
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

## 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?

• What about dividing complex numbers?
• 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!
• 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.
• 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.
• 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
• strange, i created this account myself cause i wanted to learn meths
• 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