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I am trying to work on this old qual exam.

Here is the question:

Find the number of roots (counting multiplicities) of the function $$f(z)=\cos(z)-1 + \frac{z^2}{2}$$ inside the domain $\vert z \vert <1$.

My work: I first thought of Rouché's theorem. But then I figured that $f(z)=z^4\left(\frac{1}{4!}-\frac{z^2}{6!}+\cdots\right)$. So $f(z)=z^4 g(z)$ for some analytic function $g(z)$ such that $g(0)\neq 0$. And then I used the argument principle to conclude that the number of zeroes is $4$. Is this correct?

Also, how do I know for sure that there are no other zeroes of $g$ inside the unit disk centered at $0$. Any hints? Thanks

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I'm new here. So I'm not really sure how to do that. How do I improve my accept rate? –  Jack Dawkins Jan 9 '13 at 19:31
Click on your username to see your profile. Look back at your previous questions that people submitted answers for. If you find any of the responses helpful, I suggest up voting them by clicking on the up arrow next to the number at the top left of their answer. If one particular submission answers your question "best", accept it by clicking on the check mark underneath the up/down vote button. –  JohnD Jan 9 '13 at 19:33
Thanks. Just trying to help you get the most out of the site. –  JohnD Jan 9 '13 at 20:01
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1 Answer 1

up vote 3 down vote accepted

Since $f$ is analytic in the domain $D:=\{z:|z|<1\}$, the Argument Principle says that the number of zeros of $f$ in $D$ is given by $${1\over 2\pi i}\int_D {f'(z)\over f(z)}\,dz,$$ which we will compute via the Residue Theorem.

First, expanding about $z=0$, $${f'(z)\over f(z)}=\frac{4}{z}-\frac{z}{15}+\frac{z^3}{6300}+\frac{z^5}{189000}+\cdots,$$ so by the Residue Theorem, $${1\over 2\pi i}\int_D {f'(z)\over f(z)}\,dz=\text{Res}(f'(z)/f(z),0)=4.$$

Hence, $f(z)$ has 4 zeros (counting multiplicities) in $D$.

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I still don't see how the last integral is equal to $8\pi i$. Can you please explain a little more?. Thanks. –  Jack Dawkins Jan 9 '13 at 23:37
The earlier integral I computed with Mathematica without much thought and got $4$. Perhaps the revision is clearer. –  JohnD Jan 10 '13 at 4:36
Yes, it makes sense now. Plus I was actually able to do this with Rouche's theorem as well. It never hurts to know two methods. Like one of my Professors once said, if you can prove it in two ways, it must be really really true. :).Thanks!! –  Jack Dawkins Jan 10 '13 at 7:28
Why do you only take the residue at $0$ to calculate the integral? Don't all of the zeroes of $f$ need to be considered? –  Bartek Feb 7 '13 at 0:48
@user54755 How did you solve this using Rouche's theorem? –  Arpan Dutta Aug 11 '13 at 5:24
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