Solving for z in e^{e^x} = 1: Unpacking the Question

[moxy]

Homework Statement

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

The Attempt at a Solution

This equation doesn't have any solutions, does it? e^w = 1 only when w = 0. w in this case is e^z, which is never equal to zero. I'm incredibly confused about the second part of the question... I assume $z_m$ and $z_n$ 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?

 

[susskind_leon]

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

Cheers

 

[jackmell]

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

$$\sqrt{z}$$

or:

$$\log(z)$$

Now:

$$\log(z)=\ln|z|+i(\theta+2n\pi)$$

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:

$$e^z=\log(1)$$

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

$$z=\ln(2n\pi)+i(\pi/2+2k\pi)$$

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

Just take one infinite set for now:

$$z=\ln(2\pi)+i(\pi/2+2k\pi)$$

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?