SOLUTION: Hello. My question is {{{i^-35}}}. This is a tough one for me, because i^2 is -1. Then you could multiply -1 17 times, which is still -1. Then, you have an i left over, and all

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: Hello. My question is {{{i^-35}}}. This is a tough one for me, because i^2 is -1. Then you could multiply -1 17 times, which is still -1. Then, you have an i left over, and all      Log On


   

Question 477116: Hello. My question is i%5E-35. This is a tough one for me, because i^2 is -1. Then you could multiply -1 17 times, which is still -1. Then, you have an i left over, and all of this is over 1 I think because you are raising a negative exponent. So 1/-i? However, my calculator says that it is 5E-13+i. Any help would be greatly appreciated.
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
i%5E-35
= 1%2Fi%5E35
= 1%2Fi%5E3
= 1/-i
= i