Question

How do you Simplify this

Answer 1

First, recall some rules of exponents.


`a^m * a^n = a^(m + n)`


`(a^m)/(a^n) = a^(m - n)`


`-a^n = - (a^n)`


`a^(-n) = (1)/(a^n)`

Here is the solution.


`(-i^(-44) \cdot i^(35))/(i^8) =`


`= -(i^(-9) )/(i^8)`


`= -i^(-17)`


`= -(1)/(i^(17))`


`= -(1)/(i)`

(For an explanation of how
`i^(17)` became
`i`,
see below the final answer.)


`= -(1)/(i) * (i)/(i)`


`= - (i)/(i^2)`


`= (-i)/(-1)`


`= i`

Final answer:
`i`



Integer powers of i.


`i^0 = 1`


`i^1 = i`


`i^2 = -1`


`i^3 = -1 * i = -i`


`i^4 = i^3 * i = -i * i = 1` Notice that
`i^4 = i^0`

Now the pattern is complete, and for each consecutive integer power, you get the same pattern of results. Let n be a positive integer multiple of 4.
Then,
i^n = 1
i^(n + 1) = i
i^(n + 2) = -1
i^(n + 3) = -i

After power n + 3, power n + 4 is again an multiple of 4, and you start again with 1. Notice that integer 17 is 16 + 1, so it is 1 more than a multiple of 4, so i^17 is the same as i^(n + 1) = i.