2
$\begingroup$

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

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

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

Classifications:

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.

$\endgroup$
  • $\begingroup$ 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
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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 $z^*=\infty$ ?

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
2
$\begingroup$

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

both

Addendum

@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];
$\endgroup$
  • $\begingroup$ These are Mathematica graphics, right? Which function did you use? $\endgroup$ – nilo de roock Mar 20 at 15:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.