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
Best Answer
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 $D$.