Finding Largest r in the Complex Exponential Function Mapping D(z0,r)

[Raziel2701]

Homework Statement

According to the Inverse Function Theorem, for every z_0 \in C there exists r > 0 such that the exponential function f(z) = e^z maps D(z0; r) invertibly to an open set U = f(D(z_0; r)). (a) Find the largest value of r for which this statement holds, and (b) determine the corresponding open set U in case z0 = 0: For part (b) use Mathematica's ParametricPlot or ContourPlot functions..

The Attempt at a Solution

I can't find anything in my book that remotely tells me how to go about finding r. I've been rather bad at this class and I would appreciate if someone could point me out to better books on complex analysis that are not Marsden's Basic Complex Analysis. I feel like I get very hard questions and they are usually like this one and I find the book to be quite hard to decipher and understand.

I would like to know first how this mapping is determined. The function e^z takes a disk of some radius centered at z0 and maps it into an open set. I don't know what to make of it, is that a very generic description or something? In any case, what kind of limitations am I supposed to be thinking, what kind of conditions should there be to figure out what the largest r would be and ultimately, how do you find this r?

 

[ystael]

Some good and readable introductory texts are Conway, Functions of one complex variable I; Stein and Shakarchi, Complex analysis; Ahlfors, Complex analysis (the classic). Many people also like Needham, Visual complex analysis, but I've never used it myself.

As for what's going on here: you are asked to find the biggest disc which the exponential function maps invertibly onto an open set. Do you understand why the image of a disc under \exp is an open set? Do you understand in what ways the mapping could fail to be invertible?

 

[Raziel2701]

The image of a disc under the exponential function is an open set because the log(z) takes strips of intervals y0i to y0i + 2pi right?

So because of this range of 2pi, the greatest the radius could be is pi, otherwise we'd be jumping to another branch of log correct?

 

[ystael]

You have reached the right answer using the right intuition, but your words don't form a coherent argument.

The image of any open set under \exp is an open set because of the open mapping theorem, which says that any analytic function with nonvanishing derivative maps open sets onto open sets. The derivative of \exp is \exp, which is never zero.

As for the radius of the disc, the mapping can fail to be invertible only by duplicating a value (failing to be injective). That is, \exp restricted to D(z_0; r) is not an invertible mapping exactly when there are z \neq z' \in D(z_0; r) such that \exp z = \exp z'. Write out this condition in terms of the real and imaginary parts of z and z', and you should be able to come up with a proof.

"Branches" are a concept that often causes more harm than good to beginning complex analysis students.