Question

how do you solve i to the negative 7 power?

Answer 1

the answer is i

Answer 2

but how do you solve it? i to the negative 7 power

Answer 3

i is defined as √(-1)

Therefore, it's in cycles of 4:

i = √(-1)
i² = -1
i³ = - √(-1) (or just -i)
i^4 = 1

So first, let's get the negative exponent out of there:

i^(-7) is the same as 1/i^7

Answer 4

Anything to the negative power is the same as putting it in the denominator and using the positive exponent:

i^0 = 1
i^1 = i
i^2 = -1
i^3 = -i

Then it repeats this pattern for every four powers:
i^4 = 1
i^5 = i
i^6 = -1
i^7 = -i

Answer 5

rewrite i in polar coordinates, exp(j*pi/2) and raise that to the -7th which becomes exp(-j7pi/2), then using Euler's Identity rewrite it as cos(7pi/2) - i*sin(7pi/2), which gives you +i

Answer 6

1/-i is i.

Answer 7

thanks for your help