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The complex exponential function maps D(z0,r) to an open set. Need to find largest r.

  • Thread starter Raziel2701
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Homework Statement


According to the Inverse Function Theorem, for every [tex]z_0 \in C[/tex] there exists r > 0 such that the exponential function [tex]f(z) = e^z[/tex] maps D(z0; r) invertibly to an open set [tex]U = f(D(z_0; r))[/tex]. (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?
 

Answers and Replies

  • #2
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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 [tex]\exp[/tex] is an open set? Do you understand in what ways the mapping could fail to be invertible?
 
  • #3
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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?
 
  • #4
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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 [tex]\exp[/tex] 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 [tex]\exp[/tex] is [tex]\exp[/tex], 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, [tex]\exp[/tex] restricted to [tex]D(z_0; r)[/tex] is not an invertible mapping exactly when there are [tex]z \neq z' \in D(z_0; r)[/tex] such that [tex]\exp z = \exp z'[/tex]. Write out this condition in terms of the real and imaginary parts of [tex]z[/tex] and [tex]z'[/tex], 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.
 

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