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In an exercise, I'm asked to classify the singularities of these functions:

$\qquad i) f(z)=\frac{1}{(z-1)^2} \qquad ii)f(z)=\frac{1-\cos z}{z^2} \qquad iii) f(z)=\frac{z^2-1}{z-1}$

I don't know why but I'm afraid I'm missing something in relation to Laurent series, poles, isolated singularities, removable singularities, etc. This is what I've done so far, and I'd be grateful if anyone could take a look and correct me if I'm wrong.

$i)$ Since $f$ is holomorphic in $D'(1,r)$ for any $r$:

$\qquad \lim_{z\to 1}(z-1)^2\frac{1}{(z-1)^2}=1$

so the limit is not zero, and then $1$ would be a pole of order $2$.

$ii)$ Here I tried writing the Laurent series' for $f$:

$\qquad \frac{1-\cos z}{z^2}=\frac{1}{z^2}-\frac{1}{z^2}\sum_{n=0}^{\infty}\frac{(-1)^n z^{2n}}{(2n+1)!}=\frac{1}{z^2}-\frac{1}{z^2}+\frac{1}{2!}-\frac{z^2}{4!}+\frac{z^4}{6!}-\cdots=\sum_{n=0}^\infty(-1)^n\frac{z^{2n}}{(2n+2)!}$

And since the series doesn't have negative powers of $z$, $f$ has a removable singularity at $0$.

$iii)$ Here I thought I could apply Riemann's theorem on removable singularities, and because

$\qquad \lim_{z\to1}(z-1)\frac{z^2-1}{z-1}=1-1=0$

then $f$ has a removable singularity at $1$.

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  • $\begingroup$ Unless the question also includes the point at infinity, this looks absolutely fine. $\endgroup$ – Chappers Jan 30 '16 at 20:49
  • $\begingroup$ Your analysis is solid! Well done. $\endgroup$ – Mark Viola Jan 30 '16 at 22:58
  • $\begingroup$ @Chappers, if it does include the point at infinity, what should I do? $\endgroup$ – ALannister Apr 26 '16 at 3:49

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