Review complex number addition, subtraction, and multiplication.
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:
Express your answer in the form .
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 .
Your answer should be a complex number in the form where and are real numbers.
Want to join the conversation?
- What about dividing complex numbers?(27 votes)
- 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!(28 votes)
- Evaluate 1+i+i^2+i^3+...+I,^1000(4 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.(5 votes)
- When is a practical situation to apply imaginary numbers?(7 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.(10 votes)
- How do you get Multiplication on the table above?(6 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.(7 votes)
- What do you do when a number is put to the power of i?
For example 2^i(4 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(9 votes)
- parents check your grade and send you to khan academy😰(6 votes)
- I think there's an error with the Multiplication formula (if I'd call it that)
It should be (a1a2+b1b2)...
correct me if I'm wrong though(2 votes)
- Sorry, the formula is correct. (b1)i*(b2i) = b1b2i^2
Since i^2 = -1, the b1b2 switches negative.
Hope this helps.(6 votes)
- get harder but I am fine(4 votes)