Find the winding number and number of zeros of certain function about $|z|=2$.

Question

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?

Answer 1

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:

$n(f,C)=\frac{1}{2\pi i}\int_0^{2\pi}3i(1+O(1/R))dt=3+O(1/R) \to 3$

Answer 2

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 $$ \frac{1}{2πi}\int_{|z|=2}\frac{f'(z)}{f(z)}dz=N_f-P_f=5-2=3$$