- #1
Bacat
- 151
- 1
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.
[tex]f(z) = \pi Cot(z\pi)[/tex]
Homework Equations
Isolated Singularities:
(1) Removable Singularity: [tex]|f(z)|[/tex] remainds bounded as [tex]z \to z_0\;[/tex]
(2) Pole: [tex]\lim_{z\to z_0} |f(z)| = \infty\;[/tex]
(3) Essential Singularity: Neither (1) or (2).
Order of a Pole:
Consider [tex]\frac{1}{f(z)}[/tex] and see how fast it approaches 0.
The Attempt at a Solution
I found isolated singularities at [tex]z=0,\;z=1[/tex].
[tex]z=0[/tex]:
[tex]\lim_{z\to 0} |f(z)| = \infty[/tex]
[tex]z=1[/tex]:
[tex]\lim_{z\to 1} |f(z)| = \infty[/tex]
Therefore, these are both poles. But I'm not sure how to find the order of the poles for a transcendental function.
Looking at [tex]\frac{1}{f(z)}[/tex]...
[tex]\frac{1}{Cot(z\pi)} = Tan(z\pi) = \frac{Sin(z\pi)}{Cos(z\pi)}[/tex]
When [tex]z\to 0[/tex], [tex]\frac{Sin(z\pi)}{Cos(z\pi)} \to 0[/tex]. To see how fast, I try an expansion of Sin...
[tex]Sin(z\pi) = z\pi - \frac{z^3 \pi^3}{6} + \frac{z^5 \pi^5}{120} - ...[/tex]
Then we have zero on the first term already. Does this mean that the z=1 pole has order zero?. 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 [tex]z=1[/tex].
[tex]\frac{1}{Cot(z\pi)} = Tan(z\pi) = \frac{Sin(z\pi)}{Cos(z\pi)}[/tex]
When [tex]z\to 1[/tex], [tex]\frac{Sin(z\pi)}{Cos(z\pi)} \to 0[/tex].
To see how fast, I try an expansion of Sin...
[tex]Sin(z\pi) = z\pi - \frac{z^3 \pi^3}{6} + \frac{z^5 \pi^5}{120} - ...[/tex]
[tex]z=1[/tex] gives [tex]Sin(z\pi) = \pi - \frac{\pi^3}{6} - ...[/tex]
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?