Imaginary Numbers:
i is an imaginary number. The general formula is a + bi. (Complex number)
i^2 = -1
You could also say: sqrt(-1) = 1
When you are graphing these kinds of equations, you graph it in a special way. The x-axis value is indicated by a, while the y-axis value is indicated by b.
9i + 4i = 13i
You combine them together when doing this. Multiplication looks like this:
9i * 2i = 18i^2
Here's simplifying:
4i + 6i^2
Remember that i^2 is equal to -1, so 6 x -1 = -6
4i + -6 = 4i - 6
Basically you just simplify as much as possible like all other equations. You can also solve things like this:
x^2 + 1 = 0
What's x? Since we know that any positive and negative number square is ALWAYS positive, it HAS to be imaginary. We then do the problem like normal:
x^2 = -1
x = i because i^2 = -1
Here's some info on distribution:
5i(2 + 6i)
5i(2 + 6i) = 10i + 30i^2 =
10i + -30 =
10i - 30 =
-30 + 10i (a + bi form of the answer where a = -30 and bi = 10i)
Powers:
When doing powers, you can do fractions too. These fractions are the same as taking roots from numbers.
x to the power of 1/2 is the same thing as the square root of x
x to the power of 1/3 is the same thing as the cube root of x
Here are some examples off the top of my head:
4^(1/2) = 2
8^(1/3) = 2
32^(1/5) = 2
Basically, you can reverse all of those. The answer to the power of the denominator of the fraction gives the original number.
2^2 = 4
2^3 = 8
2 ^ 5 = 32
I can also get into graphing exponential functions/decay and logarithms if you need since those topics are related. Logarithms take it a step further though. If you want to know if you don't already just post. If you haven't covered Logs then I bet you will soon because that usually comes after messing around with powers and roots. (Although you might do some Synthetic/long polynomial division and find roots of polynomials first)
I didn't give the best explanation in the world on either of these topics, so just let me know if further detail is needed and what you don't understand. (I love math
)
EDIT: Oh and here's how to solve -3^(1/2):
-3 ^ (1/2) = sqrt(-3)
sqrt(-1) = i
sqrt(-3) = 3i
Why?
Well 3i^2 is the same thing as -3, so take the square root of that and you get sqrt(-3)