Find the winding number and number of zeros of certain function about $|z|=2$. I have the function $f(z)=z^3+\frac{1}{(z-1)^2}$ and I am asked to find the winding number about $C:=\{|z|=2\}$ and then the number of zeros inside $C$.
I know that the winding number is:
$$
n(f,C)=\frac{1}{2\pi i}\int_C \frac{f'(z)}{f(z)}dz,
$$
but this gives the integral:
$$
\frac{1}{2\pi i}\int_C \frac{3z^2(z-1)^3 -2}{(z-1)(z^3(z-1)^2+1)}dz
$$
which is not very workable, even with the residue theorem (unless I am mistaken). Instead, if I write $C$ as $2e^{2 \pi i \theta}$ for $\theta \in [0,1]$, I can re-examine $f$ as:
$$
8e^{6\pi i \theta} + \frac{1}{(2e^{2\pi i \theta}-1)^2}.
$$ The fraction on the right is at its largest (in terms of modulus) when $\theta = 0$ and its smallest when $\theta = 1/2$. This leads me to believe that, since the left term is much larger, the curve will wind three times around.
How can I make this more rigorous?
Furthermore, by the argument principle, I get that $n(f,C)=\#\text{zeros of }f-\#\text{poles of }f$. Since $f$ has (counting with multiplicity) $2$ poles inside $C$, this would give me that $f$ has $5$ zeros inside $C$, but this seems odd.
Is this correct?

The zeroes of $f$ are the zeroes of the polynomial $P(z)=z^3(z-1)^2+1$ and for $|z| \ge 2$, one has $|P(z)| \ge 7$ by trivial majorizations so all the 5 zeroes of $P$ are inside $C$, hence $f$ has indeed $5$ zeroes there.
To compute the integral one uses the above observation that all the zeroes of the denominator are inside $C$ so by Cauchy one can move $C$ to infinity and the integral stays same - in other words:
$n(f,C)=\frac{1}{2\pi i}\int_C \frac{f'(z)}{f(z)}dz=\frac{1}{2\pi i}\int_{|z|=R} \frac{f'(z)}{f(z)}dz, R\ge 2$
But then only the ratio of the leading terms matters, so one gets by using $z=Re^{it}, dz=izdt$, dividing by $z^6$ both numerator and denominator, estimating trivially the other terms and taking $R \to \infty$ that:
$n(f,C)=\frac{1}{2\pi i}\int_0^{2\pi}3i(1+O(1/R))dt=3+O(1/R) \to 3$
