Let $f$ and $g$ be differentiable on a domain $D$ and suppose that $\gamma$ is a simple closed contour whose inside is contained in D.
If $|f(z)-g(z)|<|f(z)|$ for all $z$ on $\gamma$, then $f$ and $g$ have the same number of zeros inside $\gamma$ (counted including their order).
I was reading an example of application of Rouche's Theorem, where Rouche's theorem was used to show that the polynomial $p(z)=z^5+10z-3$ has $1$ zero in $\{z:\mathbb{C}:|z|<1\}$.
What was done:
Let $f(z)=10z-3$ and $g(z)=z^5+10z-3$. Let $|z|=1$. Then we get that:
$|f(z)-g(z)|\\=|-z^5|= |z^5|=1
Hence $p(z)$ has 1 zero inside $\{z:\mathbb{C}:|z|<1\}$.
How did they get that $1
Secondly, why did they choose $f(z)=10z-3$? Can I choose $f(z)=10z$ and is what I did below correct?
$|f(z)-g(z)|\le|f(z)|\\|-z^5+3|\le10|z|\\|z|^5+|3|\le10|z|\\4\le10$
Hence p(z) has 1 zero inside $\{z:\mathbb{C}:|z|<1\}$.