Bacat
Mar13-09, 04:51 PM
1. The problem statement, all variables and given/known data
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)
2. Relevant 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.
3. 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.
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?. 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?
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)
2. Relevant 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.
3. 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.
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?. 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?