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I want to find how many zero's does $z^{10} + 9ze^{z+1}-8$ have in the open unit disc. Do I need to apply Rouché's Theorem twice?

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Once will be enough. Hint: show $|9 z e^{z+1}| > |z^{10} - 8|$ on the unit circle. – Robert Israel Aug 7 '11 at 7:35
Thx a lot, I got it! – user786 Aug 7 '11 at 8:31
@robert: It would be nice if you make that comment, into an answer so that this question doesn't remain in the unanswered list. – user9413 Aug 7 '11 at 8:40
Or perhaps, the OP can can answer the question himself, since now he has got the idea for solving the problem. – user9413 Aug 7 '11 at 8:40
Ok, I'll answer the question. – user786 Aug 7 '11 at 8:51

1 Answer

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Following Robert's hint, let $g(z)= 9ze^{z+1}$. Then note that on $S^1$ we have $|f(z)-g(z)|=|z^{10}-8|=7$. But $|g(z)|=|9ze^{z+1}|=9|e^{z+1}|$. Now observe that if $z\in S^1$ then $|Re(z+1)|\geq 0$ so that $|e^{z+1}|=e^{Re(z+1)} \geq 1$. So we always have on $S^1$ $|e^{z+1}|> \frac{7}{9}$. By Rouché's Theorem, $f(z)=z^{10}+9ze^{z+1}-8$ has the same number of zero's as $g(z)= 9ze^{z+1}$, that is none.

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Close, but $7\le |z^{10}-8|\le 9$ on the unit circle. However, $|9z e^{z+1}|\ge 9$; the only point on the unit circle at which $|9z e^{z+1}|=9$ is $z=-1$, and $|z^{10}-8|=7$ at $z=-1$. Thus, $|f(z)-g(z)|<|g(z)|$ for all $z$ on the unit disk. One other point is that $9z e^{z+1}$ has one zero in the unit disk, namely $z=0$. – robjohn♦ Aug 7 '11 at 9:59
Thank you for the correction! – user786 Aug 7 '11 at 18:28

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