I am having trouble understanding the computational part of the argument principle. I have a question, let $f(z) = e^z+10z^3-z+\frac{1}{z-4}$ and find the number of zeros including multiplicity, inside the circle $|z|=1$.
I understand how to prove the theorem but am just having trouble applying it. Any help would be appreciated!
If $|z|=1$, then $|10z^3|=10$ and$$\left|e^z-z+\frac1{z-4}\right|\leqslant|e^z|+|z|+\frac1{|z-4|}<10=|10z^3|.$$Therefore, by Rouché's theorem, $f(z)$ has as many zeros inside the circle $|z|=1$ as $10z^3$, that is it has $3$ zeros.