Solving a Puzzling Equation: -e-x + e+x/2i

[struggles]

Homework Statement

So this is a really simple problem and i know I'm missing something really obvious but i just can't spot it.

Homework Equations

The Attempt at a Solution

so in the second part above I get :
e^-x - e^x/2i. However I don't get the next bit as the 2i on the denominator is now factored into the top equation to give -ie^-x + ie^x/2. I don't get how the 2 equations are equivalent and what they've done to get the i from the denominator to the numerator. If anyone could explain how this happened that'd be great!

 

[fresh_42]

What is i?

 

[axmls]

Do you know what $1/i$ is?

 

[struggles]

Do you know what $1/i$ is?

i^-1?

 

[fresh_42]

i^-1?

First answer my question and then use it to answer struggles'. $i^{-1}$ is simply another form of $\frac{1}{i}$ which doesn't really help.

 

[robphy]

Suppose you didn't know what $i$ is
but it seems that (from your second equals sign) $\frac {1}{i}= -i$... what must $i$ satisfy?

 

[fresh_42]

Suppose you didn't know what $i$ is
but it seems that (from your second equals sign) $\frac {1}{i}= -i$... what must $i$ satisfy?

How is it defined? What stands $i$ for? There must have been some explanation for it.
Remark: Sorry, for confusing your names, struggles and axmls!

 

[axmls]

Though it isn't obvious that this is the way to find this, here's a hint:
$$\frac{1}{i} = \frac{1}{i} \frac{i}{i}$$

 

[struggles]

so i = √-1 so i² = -1. so 1/i must equal 1/(-1)^1/2 = (-1)^-1/2? is that what you were getting at fresh-_42?
So axmls 1/i = i/i^2 = -i! Thanks both of you!