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can anybody show me how to use Cauchy's Theorem to show that for sufficiently large R, if $\gamma$ is the circle $|z|$=R oriented counter clockwise, then $\int_\gamma \frac{p'(z)}{p(z)} dz=2\pi i n$. Thanks a lot.

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  • $\begingroup$ What are $p$ and $n$? $\endgroup$
    – agha
    Feb 16, 2015 at 21:39
  • $\begingroup$ I suggest you pick up any introductory book on complex analysis, free, bought or borrowed, and look at the sections on winding numbers and the residue theorem. $\endgroup$
    – davidlowryduda
    Feb 16, 2015 at 21:39
  • $\begingroup$ I just learn about winding number, but not residue. I am not quite sure how to use the winding number part. $\endgroup$
    – Lila
    Feb 16, 2015 at 21:42
  • $\begingroup$ p is n polynomial. And n probably just the degree of polynomial? $\endgroup$
    – Lila
    Feb 16, 2015 at 21:43
  • $\begingroup$ Are you trying to prove the Fundamental Theorem of Algebra with this? $\endgroup$ Feb 16, 2015 at 21:57

1 Answer 1

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Note that if $$p(z) = C\prod_{q=1}^n (z-\rho_q)$$ then $$p'(z) = C\prod_{q=1}^n (z-\rho_q)\sum_{q=1}^n \frac{1}{z-\rho_q}$$ so that $$\frac{p'(z)}{p(z)} = \sum_{q=1}^n \frac{1}{z-\rho_q}.$$ Hence $$\mathrm{Res}_{z=\rho_q} \frac{p'(z)}{p(z)} = 1$$ and $$\int_{|z|=R} \frac{p'(z)}{p(z)} \;dz = 2\pi i n.$$

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  • $\begingroup$ Hi, thanks Marko, the thing is I need to use Cauchy's Theorem instead of FTA. $\endgroup$
    – Lila
    Feb 16, 2015 at 21:58
  • $\begingroup$ The Cauchy Residue Theorem is used on the next to last line. $\endgroup$ Feb 16, 2015 at 22:00
  • $\begingroup$ I have not learn Residue Theorem. I only learn the winding number part. $\endgroup$
    – Lila
    Feb 16, 2015 at 22:02
  • $\begingroup$ I also have learn the Cauchy integral Theorem. $\endgroup$
    – Lila
    Feb 16, 2015 at 22:03
  • $\begingroup$ Let's see what others have to say. If I have some basic misunderstanding here I will remove my answer. $\endgroup$ Feb 16, 2015 at 22:04

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