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I need to know if I'm doing this exercise correctly:

  • $f(z)$ has only one singularity, which is at $z=0$, and it's a pole of order 7
  • $f(z)=-f(-z)$
  • $f(z)$ is analytic in $z=\infty$

  • $g(z)$ is analytic everywhere except of $z=\infty$, where it has a pole of order 7

I need to calculate $Res[f.g]$ in $z=0$

I did this:

$f(z)$ has a pole of order 7 in $z=0$, so I can write:

$$f(z)=\sum_{n=-7}^{\infty} C_n.z^n$$

$g(z)$ is not analytic in $z=\infty$, so $g(\frac{1}{z})$ is not analytic in $z=0$. So:

$$g(\frac{1}{z})=\sum_{n=-7}^{\infty} D_n \frac{1}{z^n}$$

I know that the Residue of any function is $Res[h(z), z=z_0]=a_{-1}$, where $a_{-1}$ is the $n=-1$ in the series.

So, I say that:

$$f(z).g(\frac{1}{z})=\sum_{n=-7}^{\infty}C_n.z^n.D_n.\frac{1}{z^n}$$

So (evaluating the series at $n=-1$):

$$Res[f\cdot g]=C_{-1}.D_{-1}$$

Is that OK? I didn't use that $f(z)$ is odd and I didn't use that $f(z)$ is analitic in $z=\infty$

Thanks everyone!

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  • $\begingroup$ For starters, you're not multiplying series correctly. Think about how you multiply two polynomials :) $\endgroup$ – Ted Shifrin Feb 25 '15 at 1:19
  • $\begingroup$ In fact, multiplying polynomials is exactly what you need: $g(z)$ is a polynomial in $z$ and $f(z)$ is a polynomial in $1/z$. $\endgroup$ – Robert Israel Feb 25 '15 at 1:21
  • $\begingroup$ I tried to do it like that, but I can't see where is $n=-1$ if I multiply 2 polynomials $\endgroup$ – Unnamed Feb 25 '15 at 1:30
  • $\begingroup$ Think about how you get the coefficient of, say, $z^3$ if you multiply $(1+2z+z^2-5z^3+z^4)(1/z+2-z+z^2+6z^3+z^5)$. $\endgroup$ – Ted Shifrin Feb 25 '15 at 1:32

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