I came across the following problem:
Find the number of zeros of $P(z)=z^4+4z^3+6z^2-4z+4$ in the circle $|z+1|<1$. I know that I can't apply Rouche's Theorem directly since we don't have a condition of the form $|z|<M$ and thus bounding the polynomial is hard. I also tried applying the Cauchy's Argument Principle but I couldn't compute the integral $$\frac{1}{2\pi i}\int_{|z+1|<1} \frac{P'(z)}{P(z)}=\text{Ind}_{f \circ \gamma}(0),$$ where $\gamma:=|z+1|<1$.
Also, here's a solution I thought: $$ P(z)=(z+1)^4-8(z+1)+11 $$ and in $|z+1|=1$, the term $11$ dominates, so $$ |P(z)-11| \leq |(z+1)^4-8(z+1)| \leq |z+1|^4+|8(z+1)| =9<11, $$ so by Rouché's theorem the polynomial doesn't have any roots in $|z+1|<1$.
Is this correct?