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I'm new with these complex contour integrals and I wanted to know if my approach to this one is correct. The integral is $$\oint_\gamma \frac{\sin z}{z^3-z}\,\mathrm{d}z$$ where $\gamma(t)=2e^{it}, t\in[0,2\pi]$. First, the integrand $f(z)$ has three singularities in the region enclosed by $\gamma$:

  • $z=0$ that results a removable singularity
  • $z=1$, pole of order $1$
  • $z=-1$, pole of order $1$

Since the Residue Theorem involves only poles, I computed $\mathrm{Res}f|_{-1}=-\frac{\sin 1}2$ and $\mathrm{Res}f|_1=\frac{\sin 1}2$. Hence $$\oint_\gamma \frac{\sin z}{z^3-z}\,\mathrm{d}z=2\pi i(\mathrm{Res}f|_{-1}+\mathrm{Res}f|_1)=0$$ I think my calculation are right but I have a doubt about the Laurent series of $f(z)$ centered in each singularity I found… Actually, I have not understood very well how to find Laurent series of a-bit-harder-than-usual meromorphic function such as the one I proposed here… Could you tell me what I miss to compute this series?

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What you did is fine. And you do not need to compute those Laurent series at all. For instance, at $1$, $\frac{\sin(z)}{z^3-z}$ is the quotient of a function which has not zero there by a function which has a simple zero there. Therefore $\frac{\sin(z)}{z^3-z}$ has a simple pole at $1$ and$$\operatorname{res}_{z=1}\left(\frac{\sin(z)}{z^3-z}\right)=\frac{\sin(1)}{3z^2-1|_{z=1}}=\frac{\sin(1)}2.$$

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  • $\begingroup$ Perfect, I'm glad I did things well… But I still wonder which are the Laurent series of the function around its singularities $\endgroup$
    – bianco
    Apr 30 '20 at 18:42
  • $\begingroup$ It's an ugly thing. For instance, the Leurent series centered at $1$ is$$\frac{\sin (1)}{2(z-1)}+\left(\frac{\cos (1)}{2}-\frac{3 \sin(1)}{4}\right)+(z-1)\left(\frac{5 \sin (1)}{8}-\frac{3 \cos (1)}{4}\right)+(z-1)^2\left(\frac{19 \cos (1)}{24}-\frac{9 \sin (1)}{16}\right)+(z-1)^3\left(\frac{53 \sin (1)}{96}-\frac{13 \cos (1)}{16}\right)+(z-1)^4 \left(\frac{397 \cos (1)}{480}-\frac{35 \sin (1)}{64}\right)+\cdots$$ $\endgroup$ Apr 30 '20 at 18:52
  • $\begingroup$ How did you find it? $\endgroup$
    – bianco
    Apr 30 '20 at 20:01
  • $\begingroup$ I computed the first terms of the Taylor series of $\frac{\sin(z)}{z^2+z}$ and then I divided what I got by $z-1$. $\endgroup$ Apr 30 '20 at 20:07

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