Isolated singularity in complex plane

[Bacat]

Homework Statement

Locate each of the isolated singularities and tell whether it is a removable singularity, a pole, or an essential singularity. If removable, give the value of the function at the point. If a pole, give the order of the pole.

$$f(z) = \pi Cot(z\pi)$$

Homework Equations

Isolated Singularities:

(1) Removable Singularity: $$|f(z)|$$ remainds bounded as $$z \to z_0\;$$

(2) Pole: $$\lim_{z\to z_0} |f(z)| = \infty\;$$

(3) Essential Singularity: Neither (1) or (2).

Order of a Pole:

Consider $$\frac{1}{f(z)}$$ and see how fast it approaches 0.

The Attempt at a Solution

I found isolated singularities at $$z=0,\;z=1$$.

$$z=0$$:

$$\lim_{z\to 0} |f(z)| = \infty$$


$$z=1$$:

$$\lim_{z\to 1} |f(z)| = \infty$$

Therefore, these are both poles. But I'm not sure how to find the order of the poles for a transcendental function.[/color]

Looking at $$\frac{1}{f(z)}$$...

$$\frac{1}{Cot(z\pi)} = Tan(z\pi) = \frac{Sin(z\pi)}{Cos(z\pi)}$$

When $$z\to 0$$, $$\frac{Sin(z\pi)}{Cos(z\pi)} \to 0$$. To see how fast, I try an expansion of Sin...

$$Sin(z\pi) = z\pi - \frac{z^3 \pi^3}{6} + \frac{z^5 \pi^5}{120} - ...$$

Then we have zero on the first term already. Does this mean that the z=1 pole has order zero?[/color]. I might have the right answer for the wrong reason. I'm just not sure and the answer is not provided.


But this method doesn't seem to help when $$z=1$$.

$$\frac{1}{Cot(z\pi)} = Tan(z\pi) = \frac{Sin(z\pi)}{Cos(z\pi)}$$

When $$z\to 1$$, $$\frac{Sin(z\pi)}{Cos(z\pi)} \to 0$$.

To see how fast, I try an expansion of Sin...

$$Sin(z\pi) = z\pi - \frac{z^3 \pi^3}{6} + \frac{z^5 \pi^5}{120} - ...$$

$$z=1$$ gives $$Sin(z\pi) = \pi - \frac{\pi^3}{6} - ...$$

It does eventually go to zero, but I'm not sure how to determine how many terms it takes. How do I calculate the order of this pole?

 

[bob1182006]

I think you need to first convert Cot into it's Laurent Series representation and then look at that to see where the singularities are.

Also, cot goes to infinity for any value of n*pi not just 0 and 1*pi. But the Laurent series representation will show you this.

 

[HallsofIvy]

There are an infinite number of poles not just at 0 and 1! And deciding how fast 1/f goes to 0, look at Taylor's series expansions for sine and cosine.