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?

You can simply compute the numbers of poles of the given f,clearly are 2 including the order and are isnide the contour.Now for the roots of the given,look at the nominator and consider the functions $h(z)=1-2{z^4}$ and $g(z)={z^5}+{z^3}$ aply Rouche's theorem for the contour $|z|=2$ and you conclude that nominator$(f)$ and g have the name number of zeros inside the contour.The roots of g are 0,i,-i with order 3,1,1 and all are isnide the circle.Of course outside the circle we dont have zeros and poles so by the argument principle $$ 
\frac{1}{2πi}\int_{|z|=2}\frac{f'(z)}{f(z)}dz=N_f-P_f=5-2=3$$
