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For $$f(z)= z^3+\frac{1}{(z-1)^2}$$ how many times does $f(z)$ wind around the origin, as $z$ moves along the circle $|z|=2$ counterclockwise?

I see that the answer should be $$ \frac{1}{2\pi i} \int_{|z|=2} \frac{f'(z)}{f(z)}dz = \frac{1}{2\pi i} \int_{|z|=2} \frac{3z^2(z-1)^3-2}{z^3(z-1)^3+(z-1)}dz $$ but I'm not sure how to evaluate the integral since I could not find where exactly the poles of the function integrated are. Here is what I thought:

(a) Since $f(z)$ has a pole of order 2 at $z=1$, $\dfrac{f'(z)}{f(z)}$ has a simple pole at 1 with residue $-2$.

(b) It can be shown that $f(z)$ has 3 zeros of order one inside the disk $|z|<2$, so at each of the zeros, $\dfrac{f'(z)}{f(z)}$ has a simple pole with residue 1.

Does this mean that the integral evaluated is $-2+3=1$? Any help or hint is appreciated!

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  • $\begingroup$ Argument principle and you avoid computing integrals: $f(z)=\frac{z^3(z-1)^2+1}{(z-1)^2}$. The number of winds is the number of zeros of $z^3(z-1)^2+1$ inside $|z|=2$ minus the number of zeros of $(z-1)^2$ ibidem. $\endgroup$
    – Peyton
    Aug 4 '17 at 20:00
  • $\begingroup$ @Peyton I see. Just to double check: so the answer should be $5-2=3$? $\endgroup$
    – Yuxin Wang
    Aug 4 '17 at 20:08
  • $\begingroup$ Yes, I think so. $\endgroup$
    – Peyton
    Aug 4 '17 at 20:09
  • $\begingroup$ Also, $f$ has neither poles nor zeros outside the disk with radius $2$, so you can use any radius $\geqslant 2$ to compute the winding number. For a large radius, the integrand $\frac{f'(z)}{f(z)}$ is $\frac{3}{z} + \text{ insignificant bits}$, so the winding number is $3$. $\endgroup$ Aug 4 '17 at 20:14
  • $\begingroup$ @Daniel Fischer Thank you for your comment! I will formulate it and post as the answer. $\endgroup$
    – Yuxin Wang
    Aug 4 '17 at 20:18
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Thanks to the comments above, the problem can be solved easily using the Argument Principle, i.e., the solution is the number of zeros of $$f(z)=\frac{z^3(z-1)^2+1}{(z-1)^2}$$ minus number of poles of $f(z)$ inside the disk $|z|<2$, which is clearly $5-2=3$.

Alternatively, observe that $f(z)$ has no zero or poles outside the disk $|z|<2$, so we can choose a large enough disk $|z|<R$ to compute the winding number. More specifically, when $R$ is large enough such that $ \dfrac{f'(z)}{f(z)} \sim \dfrac{3}{z}$, one easily observes that the winding number is 3.

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