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On all the complex analysis-exams written by my professor, a question of this nature always pops up:

Let $$f(z)=\frac{1+e^z}{1-e^z}$$

and $A=\{z:\Re(z)<0, \ -\pi<\Im(z)<\pi\}.$ Determine the image $f(A)$. Hint: use the fact that $f(z)=M(e^z)$ where $M(z)=\frac{1+z}{1-z}.$

I'm wondering if anyone can somehow break down a solution for problems like these in steps. Like, step 1: check this, step 2: compute this and so on. I already have the profesors solutions but they are too cryptic for me. I need this problem simplified somehow so I can understand what needs to be done.

My initial thought is simply that we have an area $A$, which is an infinite rectangular area that is to the left of the complex halfplane but bounded by $-\pi$ and $\pi$ on the imaginary axis. So if every point inside of $A$ undergoes the transformation $f(z),$ they will form a different figure, i.e the image $f(A).$

However, I don't understand how I'm supposed to do the arithmetic here and use Möbius transforms.

Any help is greatly appreciated, but please no usage of fancy math that is outside the scope of complex analysis.

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    $\begingroup$ There is a hint there. It tells you to first find $e^A$, and then do the mobius transformation $z\mapsto\frac{1-z}{1+z}$ on that set. Can you find $e^A$? $\endgroup$ – Arthur Oct 26 '18 at 14:22
  • $\begingroup$ @Arthur - Well, if $A$ is a region and not an expression, how can I then exponentiate a region? And what do you mean by "doing" a transformation? Plugging in $e^{A}$ into $M(z)?$ Then I have the same question as the first one. Prolly stupid questions but... so be it. $\endgroup$ – Parseval Oct 26 '18 at 14:25
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    $\begingroup$ By $e^A$, @Arthur means the image of $A$ under $z \mapsto e^z$. Try seeing where a general element of $A$ ends up under this map. $\endgroup$ – Sam Streeter Oct 26 '18 at 14:27
  • $\begingroup$ @Artur - $e^A:=\{z:0<|z|<1, -\pi < \arg{(z)} < \pi\} = D(0,1)$. I don't see why I need to remove $(-1,0].$ $\endgroup$ – Parseval Oct 26 '18 at 15:40
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First, determine the image of $A$ under $z \mapsto e^z$. It is fairly easy to see that this will be the interior of the unit disk (minus the part of the negative real axis within the unit disk) by writing $e^{x+iy} = e^x \cdot e^{iy}$ . Then use the fact that Möbius transformations map circles and lines to circles and lines on the boundary of this region.

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  • $\begingroup$ By interior, are you including or excluding the boundary? Also, how can I see that we need to subtract the range $(-1,0]$ from the unit disk? I don't find this obvious to see. $\endgroup$ – Parseval Oct 26 '18 at 15:28
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    $\begingroup$ @Parseval You need to remove that part because there is no point of $A$ with imaginary part $\pi$ or $-\pi$. Thus there is no point of $A$ which hits any point of $(-1,0)$. And there is no complex number that gets mapped to $0$ under the exponential map at all, so specifically there is no point of $A$ which gets mapped there. $\endgroup$ – Arthur Oct 26 '18 at 17:00
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    $\begingroup$ @Parseval As to whether the boundary should be included, what points are sent to the boundary of the unit disc by the exponential map? Are those points part of $A$? $\endgroup$ – Arthur Oct 26 '18 at 17:22

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