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$$f(z)=\frac{z^3+1}{z^2(z+1)} $$

has singularities $z=0, z=-1$ right?

How can I determine the type of singularity of this points?


1.)Removable pole (Then $f(z_0)$ is bounded, $f(z)$ has a limit if $z \to \infty$

2.) Pole of order $m \implies |f(z)|\to \infty $ as $z \to z_0$

3.) Essential singularity : not bounded, does not go to infinity.

(Another way to descibre is to look at the coefficients of the Laurent Series.

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In some classes it is not really convered, so I'm curious: are you also expected to talk about the (possible) singularity at $z=\infty$? – Antonio Vargas Jan 21 '13 at 20:35
That's the next paragraph :) – Applied mathematician Jan 21 '13 at 23:50
up vote 3 down vote accepted

A function $f$ which is holomorphic for all $z$ near $z_0 \in \mathbb{C}$ (with $z \neq z_0$) has a pole of order $m>0$ at $z=z_0$ if and only if

$$ \lim_{z \to z_0} (z-z_0)^m f(z) $$

is finite and nonzero. If $f$ has a singularity at $z=z_0$ and

$$ \lim_{z \to z_0} f(z) $$

is finite and nonzero then the singularity is removable, and vice versa.

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How to apply this to my function? – Applied mathematician Jan 21 '13 at 19:54
@Joyeuse, consider the pole at $z=0$. What $m$ must you choose to make $\lim_{z\to 0} z^m f(z)$ exist and be nonzero? – Antonio Vargas Jan 21 '13 at 19:58
that would be 0... – Applied mathematician Jan 21 '13 at 20:01
@Joyeuse no... are you telling me that $$\lim_{z\to 0} \frac{z^3+1}{z^2(z+1)}$$ is finite? – Antonio Vargas Jan 21 '13 at 20:02
Glad to help! ${}$ – Antonio Vargas Jan 21 '13 at 20:32

Decompose $f$ as $$f(z)=\dfrac{z^3+1}{z^2(z+1)}=\dfrac{(z+1)(z^2-z+1)}{z^2(z+1)}.$$ Then $f$ has at $z_0=-1$ removable singularity and pole of order $2$ at $z_1=0.$ What type of singularity is at $z^*=\infty$ ?

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How can you see that the order of $z_1=0$ is $2$ so quickly? Is a removable singularity because of the cancelation? (Always the case?) If $z$ goes to $\infty$ I think it has a limit ($f \to $1$ as $z \to \infty$). ? – Applied mathematician Jan 21 '13 at 19:58
As Antonio Vargas suggested, $\lim\limits_{z \to 0} z^2 f(z)=\lim\limits_{z \to 0} (z^2-z+1)=1$ and $\lim\limits_{z \to {-1},\;z\ne{-1}} f(z)=3$ – M. Strochyk Jan 21 '13 at 20:17
thank you sir :) – Applied mathematician Jan 21 '13 at 20:46

By applying partial fraction decomposition, we get: $$ f(z) = 1 - \dfrac{1}{z} + \dfrac{1}{z^2} $$

It's now easy to see that $z = -1$ is a removable singularity. $z^3 + 1$ can be factored as $(z + 1)(z^2 - z + 1)$ so $z + 1$ can be canceled and this removes the singularity.

For the singularity at $z = 0$, it is a pole of order two as the principal part is clearly $- \dfrac{1}{z} + \dfrac{1}{z^2}$.

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