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Course: Get ready for Precalculus > Unit 1
Lesson 5: Multiplying complex numbersComplex number operations review
Review complex number addition, subtraction, and multiplication.
Addition | ||
Subtraction | ||
Multiplication | ||
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:
Example 2: Subtracting complex numbers
When subtracting complex numbers, we simply subtract the real parts and subtract the imaginary parts. For example:
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:
Unlike regular binomial multiplication, with complex numbers we also consider the fact that .
Example 1
Example 2
Example 3
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?(45 votes)
- Evaluate 1+i+i^2+i^3+...+I,^1000(7 votes)
- 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.(11 votes)
- When is a practical situation to apply imaginary numbers?(9 votes)
- rotations like how you have i, -1, -i, and +1. Look into radio signaling or transmissions, i think it applies to quantum as well.(16 votes)
- super duper very cool(14 votes)
- How do you get Multiplication on the table above?(8 votes)
- 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.(10 votes)
- What do you do when a number is put to the power of i?
For example 2^i(5 votes)- 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(11 votes)
- get harder but I am fine(7 votes)
- It's going to get harder the father down the road you go.(6 votes)
- 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(3 votes)- Sorry, the formula is correct. (b1)i*(b2i) = b1b2i^2
Since i^2 = -1, the b1b2 switches negative.
Hope this helps.(9 votes)
- 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.(3 votes)
- 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.(5 votes)
- Can you please help with this question? :(
(2 + i)(3 - i)(1 + 2i)(1 - i)(3 + i)(2 votes)- 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.(5 votes)