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

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.

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

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.

  • $\begingroup$ How to apply this to my function? $\endgroup$ – Applied mathematician Jan 21 '13 at 19:54
  • 1
    $\begingroup$ @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? $\endgroup$ – Antonio Vargas Jan 21 '13 at 19:58
  • $\begingroup$ that would be 0... $\endgroup$ – Applied mathematician Jan 21 '13 at 20:01
  • $\begingroup$ @Joyeuse no... are you telling me that $$\lim_{z\to 0} \frac{z^3+1}{z^2(z+1)}$$ is finite? $\endgroup$ – Antonio Vargas Jan 21 '13 at 20:02
  • 1
    $\begingroup$ Glad to help! ${}$ $\endgroup$ – 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 $z^*=\infty$ ?

  • $\begingroup$ 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 $z \to \infty$). ? $\endgroup$ – Applied mathematician Jan 21 '13 at 19:58
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    $\begingroup$ As Antonio Vargas suggested, $\lim\limits_{z \to 0} z^2 f(z)=\lim\limits_{z \to 0} (z^2-z+1)=1$ and $\lim\limits_{z \to {-1},\;z\ne{-1}} f(z)=3$ $\endgroup$ – M. Strochyk Jan 21 '13 at 20:17

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 $- \dfrac{1}{z} + \dfrac{1}{z^2}$.


The exposition of @Ayman Hourieh simplifies the problem.

For visual reinforcement, conformal maps are presented where Re $z$ is blue, and Im $z$ is red. On the left is the input function $$ f(z) = \frac{z^{3} + 1}{z^{2}(z+1)} = 1 - \frac{1}{z} + \frac{1}{z^{2}} $$ On the right is the input function with the singularity suppressed $$ z^{2}f(z) = \frac{z^{3} + 1}{(z+1)} = z^{2} - z + 1 $$



@nilo de roock and others have asked how to reproduce plots in Mathematica.

z := x + I y;
f[z_] := (z^3 + 1)/(z + 1);
\[Lambda] = 1;
pts = 100;
gre = ContourPlot[
   ComplexExpand[Re[f[z]]], {x, -\[Lambda], \[Lambda]}, {y, -\[Lambda], \[Lambda]}, 
   Contours -> 15, ContourShading -> None, 
   ContourStyle -> {{Blue, Opacity[0.5]}}, PlotPoints -> pts];
  • $\begingroup$ These are Mathematica graphics, right? Which function did you use? $\endgroup$ – nilo de roock Mar 20 '19 at 15:20

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