Determine the Laurent series for $f(z)=\frac{1}{z(z+5)}$ valid in the region$ \{z: |z| < 5\}$

When i expand i got,

$f(z)= \sum_{n=0}^{\infty}\frac{(-1)^nz^{n-1}}{5^{n+1}}$. But Contour here is circle with center 0, radius 5.

But simple poles are 0, -5. O is inside given region C:$|z|<5$. Laurent series expansion pre-condition, f(z) analytical through out within C(is it on C? - i have a doubt here). So is this expansion valid???? - No i think. Pls correct me if i am wrong

  • $\begingroup$ "Laurent series expansion pre-condition, f(z) analytical through out within C"-this is not true. $\endgroup$ – Abishanka Saha Jan 14 '18 at 4:23
  • $\begingroup$ but theorem says f(z) analytic in annulus region between $C_2$ (outer circle) and $C_1$(inner circle). Here $C_1 \to 0$. So f(z) should be analytical with in $C_2$ which is C in problem. Can you pls point to mistake? $\endgroup$ – Magneto Jan 14 '18 at 4:25
  • $\begingroup$ Yes, $f(z)$ s analytic in the annulus region $0<|z|<5$, because it doesn't contain either $0$ or $5$, which are the poles of this function. $\endgroup$ – Abishanka Saha Jan 14 '18 at 4:29
  • $\begingroup$ What contour? $|z|<5$ is an open disk that is bounded by the circle $|z|=5$. $\endgroup$ – Mark Viola Jan 14 '18 at 4:31
  • 1
    $\begingroup$ Technically you are right: the expansion is valid in $0 < |z| < 5$, but not at $z=0$. However, in homework or an exam it's a good idea to answer the question that was meant, not just what was actually written. You might even get extra points for pointing out the error. $\endgroup$ – Robert Israel Jan 14 '18 at 6:00

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