How many zeros does $\sin(z)-100z^9$ have with $|z|<1$? I have to find out how many zeros $\sin(z)-100z^9$ has with $|z|<1$
My approach was to use Roché's Theorem in the following way:
Let $g(z) = \sin(z)-100z^9$ and $f(z) = -100z^9$
For $|z| = 1$ we get:
$|g(z)-f(z)| = |\sin(z)| = \sin(1) < 100 = |-100z^9| = |f(z)|$
I then concluded that $g(z)$ has $9$ zeros where $|z|=1$
I was wondering if this is the correct use of the Roché theorem or if I'm completely off.
Thank you so much for any help or suggestions.
 A: You are conclusion is right but your argument has many mistakes. Look at the statement of Rouche's Theorem again . It  states that for any two complex-valued functions f and g holomorphic inside some region
${\displaystyle K}$ with closed contour ${\displaystyle \partial K}$ if $|g(z)| < |f(z)|$ on ${\displaystyle \partial K}$, then $f$ and $f + g$ have the same number of zeros inside ${\displaystyle K}$, where each zero is counted as many times as its multiplicity.
So by how you defined it we end up with $f+g=\sin(z)-200z^{9}$.
Also $|\sin(z)|\neq \sin(1)$ for $|z|=1$. You cannot conclude say that $f(|z|)=|f(z)|$ for some function. This is not true.
For example if I take $z=i$ then $|z|=1$ but $\displaystyle |\sin(i)|=|\frac{e^{-1}-e^{1}}{2}|>1$ . You have to keep in mind that the arguments are complex and not real . $|\sin(x)|\leq 1$ is only valid for real numbers $x$. But you can in fact have $|\sin(z)|$ as large as you want.
Here's how you should do it:-
Let $f(z)=-100z^{9}$ and $g(z)=\sin(z)$
Then on $|z|=1$ ,
You have $$|g(z)|=|\sin(z)| = |\frac{e^{iz}-e^{-iz}}{2i}| = |\frac{e^{2iz}-1}{2ie^{iz}}|\leq \frac{1}{2}\bigg(\frac{|e^{2iz}|+1}{|e^{iz}|}\bigg)\\\leq \frac{|e^{2i(\cos(\theta)+i\sin(\theta))}|+1}{2|e^{i\cos(\theta)-\sin(\theta)}|}\leq \frac{|e^{-2\sin(\theta)}\cdot e^{2i\cos(\theta)}|+1}{2\cdot e^{-\sin(\theta)}}\\\leq \frac{(e^{-2\sin(\theta)}+1)}{2\cdot e^{-\sin(\theta)}}\leq \frac{e^{2}+1}{2e}$$  where $z=\cos(\theta)+i\sin(\theta)\,0\leq\theta<2\pi$ is an arbitrary point on the unit disc.
And $|f(z)|=100\,\,,\forall |z|=1$.
Now you have $|f(z)|>|g(z)|$ for all $|z|=1$ which is the boundary of the disc $|z|\leq 1 $  and hence by Rouche's Theorem you have that $f+g$ have the same number of zeroes as $f$ inside the unit disc which is $9$.
