# Reversing substitution

1. Nov 24, 2012

### pierce15

Hi guys... I'm probably missing something pretty basic here but I can't seem to figure this out. I was working on a problem recently: for the complex functions f(z)=ez and g(z)=z, find their intersections. This post is not about the problem, it is about something I noticed while tackling it (incorrectly).

Anyways, here's what I noticed: If you set these functions equal to each other, you get

ez=z

So, naturally:

z=ln(z)

From here I saw that a basic substitution was applicable, so the equation can be rewritten:

ez=ln(z)

Basically, what I have shown is that the function h(z)=ez-z has the same zeroes as the function i(z)=ez-ln(z). Now here's what's troubling me: what if the original problem that I gave you was to find the zeroes of i(z)? Originally, we obtained i(z) from h(z) by using a substitution, but is there some way that we can go in reverse from i(z) to h(z) using a "reverse substitution"? I'm sorry if this is rather unclear. Is there anything fundamental that I am missing?

Thanks a lot

2. Nov 24, 2012

### HallsofIvy

Staff Emeritus
You can do, pretty much any whacky thing you want because everything you do starts from the assumption that z is a real number such that $e^z= z$ and there is NO such number.

3. Nov 24, 2012

### pierce15

Note that we are working with complex functions. By the way, I did find an answer to this problem, right now I am just thinking about reversing the substitution

4. Nov 24, 2012

### pwsnafu

5. Nov 25, 2012

### pierce15

pwnsnafu, like I said, I already solved the equation using the labert w function. Please read my first post carefully.

6. Nov 25, 2012

### pierce15

I guess this is kind of unclear. Here is the problem stated more clearly:

Prove that the system of f(z)=ez and g(z)=ln(z) has the same solutions as the system of f(z)=ez and h(z)=z