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

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