Find Fixed Points of e^z Complex Equation

In summary, Nigel found that x=-LambertW(-1) has one solution. He also found that if x is real, then there are no fixed points.
  • #1
nigelvr
4
0

Homework Statement


How would one go about finding the fixed points of e^z, where z is complex (i.e. all z s.t. e^z = z)?

Homework Equations


Nothing.

The Attempt at a Solution


I've considered all the relevant formulas (de Moivre's forumla, power series, z = re^i*theta, ...).

For some reason, I'm just not getting the solution. I feel that this is going to be really obvious, and I'm not sure why I'm not getting it. Thanks.
 
Physics news on Phys.org
  • #2
The equation [itex]e^z=z[/itex] cannot be solved using elementary functions. So what exactly is it that you want?? Do you want to prove that there are fixed points?? Do you want an expression of them?? (Use something similar to Lambert's W function) Do you want to find a numerical approximation??
 
  • #3
I would like to find an expression for the fixed points, if they exist. If they happen to not exist, then I would like a proof of that.

It isn't really a homework question, the prof just asked us to think about finding the fixed points of e^x. The thing is though, the course doesn't assume we know any complex analysis, so I figured it could be solved using basic math (not even calculus).

Thanks for your reply.

Nigel
 
  • #4
nigelvr said:
I would like to find an expression for the fixed points, if they exist. If they happen to not exist, then I would like a proof of that.

It isn't really a homework question, the prof just asked us to think about finding the fixed points of e^x. The thing is though, the course doesn't assume we know any complex analysis, so I figured it could be solved using basic math (not even calculus).

Thanks for your reply.

Nigel

Maple 11 gets the solution as x = -LambertW(-1) =~= 0.3181315052-I*1.337235052, where I = sqrt(-1). Here, LambertW(z) is the solution of f(z)*exp(f(z))=z which is analytic at z = 0.

RGV
 
  • #5
Ray Vickson said:
Maple 11 gets the solution as x = -LambertW(-1) =~= 0.3181315052-I*1.337235052, where I = sqrt(-1). Here, LambertW(z) is the solution of f(z)*exp(f(z))=z which is analytic at z = 0.

RGV

It should be noted that -W(-1) is only one solution. Indeed, the Lambert W function is multivalued, so there are infinitely many values for -W(-1). Also note that the Lambert W function is not analytical at -1, since it has a branch cut there.
 
  • #6
micromass said:
It should be noted that -W(-1) is only one solution. Indeed, the Lambert W function is multivalued, so there are infinitely many values for -W(-1). Also note that the Lambert W function is not analytical at -1, since it has a branch cut there.

The branch point is at z_c = -exp(-1), so z = -1 is, indeed, on (one side of) the branch cut.

RGV
 
  • #7
nigelvr said:
I would like to find an expression for the fixed points, if they exist. If they happen to not exist, then I would like a proof of that.

It isn't really a homework question, the prof just asked us to think about finding the fixed points of e^x. The thing is though, the course doesn't assume we know any complex analysis, so I figured it could be solved using basic math (not even calculus).

Thanks for your reply.

Nigel

Then you should assume x is real. Think about the graph of the real functions.
 
  • #8
I really think we're missing the point with this branch-cut thing: It's arbitrary. That means I can move it and the function becomes perfectally analytic at z=-1. In fact, I would argue the function is everywhere analytic except at the branch-point [itex]z=-1/e[/itex].

Also, if I may be allowed to be complete, the OP specifically stated complex z so no real for me and to find an explicit expression for z, we get it into suitable Lambert-w form:

[tex]z=e^z[/tex]

[tex]1=ze^{-z}[/tex]

[tex]-1=-ze^{-z}[/tex]

at that point, take the W function of both sides:

[tex]-z=W(-1)[/tex]

then:

[tex]z=-W(-1)[/tex]

and keep in mind the function [itex]e^z-z[/itex] is a non-polynomial entire function which by Picard, reaches all values with at most one exception, infinitely often so that we would expect the expression [itex]e^z=z[/itex] to have an infinite number of solutions.
 
Last edited:
  • #9
I did a little exploration of this interesting problem... the first few solutions found are

0.318131505 ± i * 1.337235701
2.06227773 ± i * 7.588631178
2.653191974 ± i * 13.94920833
3.020239708 ± i * 20.27245764
3.287768612 ± i * 26.5804715
3.498515212 ± i * 32.88072148
3.672450069 ± i * 39.17644002
3.820554308 ± i * 45.4692654

... the imaginary portion increasing by approx [itex]2\pi[/itex] each time and |z| = eRe(z).
 

1. What is a fixed point in mathematics?

A fixed point is a number that remains unchanged when a certain operation is performed on it. In other words, if the number is plugged into an equation, the result will be the same number.

2. How do you find the fixed points of e^z complex equation?

To find the fixed points of e^z complex equation, you need to set the equation equal to the original variable. This will result in a quadratic equation, which can then be solved using the quadratic formula to find the fixed points.

3. Why is finding fixed points important?

Finding fixed points is important in many areas of mathematics, including complex analysis. They can provide valuable information about the behavior of a function and can help in solving equations and understanding the stability of systems.

4. Can a complex equation have multiple fixed points?

Yes, a complex equation can have multiple fixed points. In fact, most complex equations have more than one fixed point.

5. Are there any techniques for finding fixed points without solving a quadratic equation?

Yes, there are other techniques for finding fixed points, such as using iterative methods and graphical methods. These techniques can be useful in cases where solving a quadratic equation is not feasible.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
510
  • Calculus and Beyond Homework Help
Replies
7
Views
694
  • Calculus and Beyond Homework Help
Replies
6
Views
483
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
478
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
422
  • Calculus and Beyond Homework Help
Replies
12
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
938
Back
Top