# Help with simple math problem

1. Oct 17, 2005

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.

2. Oct 17, 2005

anyone help?

3. Oct 17, 2005

### James R

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

4. Oct 18, 2005

### HallsofIvy

Staff Emeritus
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.)

Last edited: Oct 18, 2005