• Support PF! Buy your school textbooks, materials and every day products Here!

Fixed points of e^z

  • Thread starter nigelvr
  • Start date
  • #1
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.
 

Answers and Replies

  • #2
22,097
3,277
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
4
0
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
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,705
1,722
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
22,097
3,277
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
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,705
1,722
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
BruceW
Homework Helper
3,611
119
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
1,796
53
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
473
13
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).
 

Related Threads for: Fixed points of e^z

  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
2K
Replies
2
Views
1K
Replies
1
Views
773
  • Last Post
Replies
3
Views
1K
Replies
3
Views
13K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
782
Top