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Show $z^4 + e^z = 0$ has a solution in $\{z \in \mathbb{C} : |z| \leq 2\}$.

I would like if in the proof the tools of algebraic topology were preferred over the other tools of analysis, complex analysis, algebra etc.

Dávid Natingga
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The tools are the same. Let $f(z)=z^4+e^z$ and let $g(z)=z^4$. Show that $\dfrac{f}{|f|}$ and $\dfrac{g}{|g|}$ are homotopic as maps $\{z: |z|=2\} \to S^1$. Note that if $h\colon S^1\to S^1$ has nonzero degree, then $h$ has a root in $D^2$. (Why?)

Ted Shifrin
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  • But this is just re-proving Rouché's theorem... – Potato May 27 '13 at 22:03
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    Yes, @Potato, whether complex analysis or differential topology or algebraic topology, it's all the same idea :) P.S. The argument principle is just a generalization of the intermediate value theorem, and differential topology puts it all in the most general context. – Ted Shifrin May 27 '13 at 22:04
  • Is there a way to see your last claim (nonzero degree implies the existence of a zero) topologically, without using the argument principle? – Potato May 27 '13 at 22:06
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    @Potato, yes, if no zeroes on the disk, then the map extends to the disk, so that gives a homotopy between the map and the constant map. (Think of the disk as a cylinder with the top circle identified to a point.) – Ted Shifrin May 27 '13 at 22:08
  • Nice! I've seen similar arguments before, but they always seem a bit tricky to me. Do you know of a good reference for topological techniques in complex analysis? Some searching brought up a book by Beardon, but it's out of print. – Potato May 27 '13 at 22:17
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    Hmm, not truly, I picked more of this up from differential topology, to be honest. But Narasimhan's and Krantz's books (http://www.amazon.com/Complex-Analysis-Variable-Raghavan-Narasimhan/dp/0817641645/ref=sr_1_1?ie=UTF8&qid=1369694488&sr=8-1&keywords=narasimhan and http://www.amazon.com/Complex-Analysis-Geometric-Mathematical-Monographs/dp/0883850354/ref=sr_1_2?s=books&ie=UTF8&qid=1369694544&sr=1-2&keywords=krantz+complex) play up the interplay with topology and differential geometry (the latter in a serious way). – Ted Shifrin May 27 '13 at 22:42
  • @TedShifrin I do not think I get your argument. How do you use the fact that the disk radius is 2? We could define a disk of a radius $\epsilon$ and there does not seem anything to fail in your argument. Of course, for a sufficiently small $\epsilon$ the equation does not have the solution. – Dávid Natingga May 28 '13 at 17:16
  • Right, David. Note that you have to make sure the homotopy between $f$ and $g$ cannot pass through $0$ (because we're projecting onto $S^1$), so you actually have to pay attention to $|f|$ and $|g|$ on the boundary circle! – Ted Shifrin May 28 '13 at 17:18