help with simple math problem

[badman]

i forgot how to do this type of problem and i i was wondering if you guys can help me out.


(1+i)^n-1=i_a im trying find i given i_a. n is the number of terms which i also have. i know u have to add the negative one to the other side. but no clue as to go from there.

[badman]

anyone help?

[James R]

Is this a complex number problem? What is i_a? Is it just a constant?

[HallsofIvy]

You won't get much help if you don't state the problem more clearly. Can I take it that i and i_a are just numbers- If you are given i_a, solve for i?
Even then is (i+1)^n-1 supposed to be (i+1)n-1 or (i+1)n-1?

In either case solving the equation is just taking a root.
If (i+1)^{n-1}= i_a then i+1= ^{n+1}\sqrt{i_a} so i= ^{n+1}\sqrt{i_a}-1.
If (i+1)^n-1= i_a then (i+1)^n= i_a+1 so
i+1= ^n\sqrt{i_a+1} and i= ^n\sqrt{i_a+1}-1.

(^n\sqrt{} is the nth root.)