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!

  • $\begingroup$ 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 $3$. $\endgroup$ – Daniel Fischer 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.