Question

"Simplify completely.Remember to put imaginary numbers in a+bi form."

5i^6

(15+7i)-(-8+13i)

3-4i/5i

2-3i/7+2i

Please explain thoroughly what you can. No pressure >_< ,I just need help .

Answer 1

So, first we need to understand how the powers of i work. The answer repeat every time you go through 4 powers. So
i^1 = sqrt(-1)
i^2 = -1
i^3 = -i
i^4 = 1
After that it repeats. So i^5 = sqrt(-1), i^6 = -1, i^7 = -i, i^8 = 1, etc.
So for the first problem, it wants 5i^6. Well, i^6 = i^4*i^2, which is (1)(-1) = -1. So then we just have 5(-1) = -5

Just like we can combine like terms with x, we can combine like terms with terms containing i. So something like 3i + 5i = 8i, theyre like terms. So for the 2nd question, you can just distribute your minus sign and combine like terms. Shouldnt be that bad.

For the 3rd question, I want to get rid of the i in the denominator. I can do this by simply multiplying top and bottom by i and then simplifying.

The 4th question makes getting i out of the bottom a bit different. Whenever I have something plus or minus i, in order to eliminate it, I need to multiply by something we call the conjugate. The conjugate is simply taking the expression and changing the sign in the middle. So if you give me 5 + 3i, the conjugate is 5-3i.
So the question is (2-3i)/(7+2i). In order to solve this, I need i out of the denominator, hence we multiply top and bottom by the conjugate of the denominator. Since the conjugate is just taking the expression and changing its sign, I will be multiplying top and bottom by 7-2i. So I would then need to simplify this:
\[\frac{ (2-3i)(7-2i) }{ (7+2i)(7-2i) }\] From there, foil out the top and foil out the bottom separately and then simplify. If you have any trouble doing any of that, feel free to ask, but hopefully you understand what you need to attempt to do and why :)