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Complex analysis, double exponential. I don't understand the question being asked.

  • Thread starter moxy
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  • #1
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Homework Statement


"For what values of z does [tex]e^{e^x} = 1[/tex] ? If [itex]z_m[/itex] and [itex]z_n[/itex] range over distinct roots of this equation, is the set of distances [itex]d(z_m, z_n)[/itex] bounded away from zero?"


The Attempt at a Solution


This equation doesn't have any solutions, does it? ew = 1 only when w = 0. w in this case is ez, which is never equal to zero. I'm incredibly confused about the second part of the question... I assume [itex]z_m[/itex] and [itex]z_n[/itex] are sequences, but that's all I've got.

At this point, I'm more interested in actually figuring out what the heck the question is asking me than I am in finding the answer. Can anyone help?
 

Answers and Replies

  • #2


No mate, there are more cases where [itex]e^z = 1[/itex]. Take for example
[tex] e^{i2\pi}=1[/tex], so [itex] z=i2\pi[/itex] is also a solution.

Cheers
 
Last edited:
  • #3
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If you're going to succeed in Complex Analysis, you have got to start thinking "multivalued" all the time like when you see:

[tex]\sqrt{z}[/tex]

or:

[tex]\log(z)[/tex]

Now:

[tex]\log(z)=\ln|z|+i(\theta+2n\pi)[/tex]

That's infinitely-valued right?

So nothing wrong with takin' logs of both sides of your expression as long as you take the infinitely-valued log of both sides:

[tex]e^z=\log(1)[/tex]

[tex]e^z=\ln|1|+i(0+2n\pi)[/tex]
[tex]e^z=2n\pi i[/tex]
Now one more time:

[tex]z=\ln(2n\pi)+i(\pi/2+2k\pi)[/tex]

That's now doubly-infinitely valued except for the purist in here.

Just take one infinite set for now:

[tex]z=\ln(2\pi)+i(\pi/2+2k\pi)[/tex]

Aren't those along a vertical line with x-cood=ln 2pi and y-coord pi/2+2kpi? How about the other ones? Those are along vertical lines too and extend to infinity. Now take the distance between arbitrary points in that set? Is that distance bounded?
 
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