Determine the change of the argument of the function $f(z)=z^2+1+\frac{5}{z}+\frac{1}{z^2}$ as z traverses the unit circle once counterclockwise.
MY attempt:
By argument principle we need to fin the amount of zeroes and poles this function has in the unit circle. It has a single pole of order 2 at $z=0$. Now for the zeroes I consider the function $z^2f(z)=z^4+z^2+5z+1$ Now by Rouche's theorem, since $|z^4+z^2|<|5z+1|$ on the unit circle, we know that $f(z)$ only one zero. Thus it traverses -1 times.
IS this correct? If not, what is the mistake and what is the correct solution. If it is correct, can you improve upon my solution.
Your proof is correct. As an alternative, you can apply “Rouché's theorem for meromorphic functions“ (see for example A question about Argument Principle/Rouche's Theorem) to $f(z)=z^2+1+\frac{5}{z}+\frac{1}{z^2}$ and $g(z)= \frac 5z$: On the unit circle is $$ |f(z) - g(z)| = |z^2+1+\frac{1}{z^2}| \le 3 < 5 = |g(z)| $$ so that $$ N_f - P_f = N_g - P_g = -1 \, . $$