imaginary numbers in math..?

Question

for example- you have i to the 7th power. so its left to i to the 3rd. now when the power is o--> 1.
whats the number when the power is the 1, 2 and 3? which goes with i, which is -1. and which is -i?

Answer 1

you start with
i
i^2=-1
i^3=-i
i^4 =1 So if you have any exponent i^n divide n into four..
The remainder can be only 0,1,2,3 . n= 4d+r So i^n=i^4d *i^r and i^4d = 1
Exe: i^127= i^3 = -i becaus 127 = 4*31+3

Answer 2

Think about the definition of i (i = sqrt(-1)). sqrt(-1) to the first power is going to be itself (as is always the case when something is raised to the first power), so i^1 is i. sqrt(-1) squared is going to negate the square root (the square root of any number squared is just going to be that number), so i^2 is -1. sqrt(-1)^3 is the same as sqrt(-1)^2 times sqrt(-1)^1, or -1*i. So sqrt(-1)^3 is -i. After the 3rd power, everything repeats.

Answer 3

Its cyclical. i**0 = 1, i**2 = -1, i**3 = -i, i**4 = 1, then you start all over. Basically what happens is that you multiply i times itself to get -1, or -1 times i to get -i or -i times i to get -i**2 = 1, etc. Draw a clock diagram and you will see the pattern.