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Reversing substitution

by piercebeatz
Tags: complex functions, substitution
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piercebeatz
#1
Nov24-12, 04:31 PM
P: 232
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
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HallsofIvy
#2
Nov24-12, 05:05 PM
Math
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Thanks
PF Gold
P: 39,363
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 [itex]e^z= z[/itex] and there is NO such number.
piercebeatz
#3
Nov24-12, 06:57 PM
P: 232
Quote Quote by HallsofIvy View Post
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 [itex]e^z= z[/itex] and there is NO such number.
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

pwsnafu
#4
Nov24-12, 11:57 PM
Sci Advisor
P: 820
Reversing substitution

This is related to the Lambert W function.
piercebeatz
#5
Nov25-12, 12:28 AM
P: 232
Quote Quote by pwsnafu View Post
This is related to the Lambert W function.
pwnsnafu, like I said, I already solved the equation using the labert w function. Please read my first post carefully.
piercebeatz
#6
Nov25-12, 01:56 PM
P: 232
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


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