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Hi can somebody help me with this: Consider the polynomial $p(z) = (z-r_1)^{a_1}(z-r_2)^{a_2}$. Assume that all roots of $p(z)$ are distinct, and let $\gamma$ be a closed rectifiable curve in $\mathbb{C}$ that does not pass through any of the roots of $p$. Find, with justification, all possible values for
$$\int_{\gamma} \frac{dz}{p(z)}.$$ I have written it as partial fraction: Let $a_1=m, a_2=n$, writing as partial fractions, we have: $$\frac{1}{p(z)}=\frac{R_1}{(z-r_1)}+\cdots+\frac{R_m}{{(z-r_1)}^{m}}+\frac{R_{m+1}}{(z-r_2)}+\cdots+\frac{R_{m+n}}{{(z-r_2)}^{n}}$$ which leads to $$\begin{aligned} &R_1(z-r_1)^{m-1}(z-r_2)^n+R_2(z-r_1)^{m-2}(z-r_2)^n+\ldots+R_m(z-r_2)^n\\ &+R_{m+1}(z-r_2)^{n-1}(z-r_1)^m+R_{m+2}(z-r_2)^{n-2}(z-r_1)^m+\ldots+R_{m+n}(z-r_1)^m=1\ldots(\ast) \end{aligned}$$ But I don't know what to do next.

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  • $\begingroup$ I don't think you need all that work, seems to me like there are 3 options, do you see why? $\endgroup$
    – dietervdf
    Commented Feb 23, 2015 at 0:25
  • $\begingroup$ Ohh? I don't see it. I just learn this thing. I did the partial fraction because I saw an example which the roots are all different. $\endgroup$
    – Lila
    Commented Feb 23, 2015 at 0:27
  • $\begingroup$ Do you know the residue-theorem? $\endgroup$
    – dietervdf
    Commented Feb 23, 2015 at 0:28
  • $\begingroup$ So basically if both roots lie outside the curve, then the integral is 0. Correct? That is 1 possibility. $\endgroup$
    – Lila
    Commented Feb 23, 2015 at 0:28
  • $\begingroup$ hmm, i meant four solutions :) $\endgroup$
    – dietervdf
    Commented Feb 23, 2015 at 0:29

1 Answer 1

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The interior of the closed curve will contain either $z_1,z_2$, both or neither. If the curve encompasses neither, the integral is zero by Cauchy. If the curve only contains $z_1$, for example, then

$$\int_{\gamma} \dfrac{1}{p(z)} \ dz = \int_{\gamma} \dfrac{1/(z-z_2)^{a_2}}{(z-z_1)^{a_1}} \ dz = \dfrac{2\pi i}{(a_1-1)!}\dfrac{d^{a_1-1}}{dz^{a_1-1}}\dfrac{1}{(z-z_2)^{a_2}}\Big{|}_{z=z_1}$$

See Cauchy's Integral Formula for more.

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  • $\begingroup$ If the curve contain both $z_1, z_2$, then what will be the values of the integral? $\endgroup$
    – Lila
    Commented Feb 23, 2015 at 0:31
  • $\begingroup$ You separate the curve $\gamma$ into $\gamma_1$, $\gamma_2$. One around $z_1$, one around $z_2$ (each small enough to not contain both) $\endgroup$
    – David P
    Commented Feb 23, 2015 at 0:33
  • $\begingroup$ Thanks a lot! I will try that then =) $\endgroup$
    – Lila
    Commented Feb 23, 2015 at 0:36

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