# Isolated Singularities: Removable Singularities and Poles

1. Jul 7, 2013

### Tsunoyukami

I'm working on a few questions for an assignment but am unsure whether my approach to this type of question is sufficient or valid. I will show my solutions to two problems that are not part of the assignment just to ensure my method is correct.

"...locate each of the isolated singularities of the given function and tell whether it is a removable singularity, a pole, or an essential singularity (case (3)). If the singularity is removable, give the value of the function at the point; if the singularity is a pole, give the order of the pole.

1. $\frac{e^{z} - 1}{z}$" (Complex Variables, 2nd edition; by Stephen D. Fisher, pg. 150)

To determine where the isolated singularity of this function occurs I look at where it is not defined (ie. where the denominator is equal to zero). This occurs when z = 0. Next, to determine which type of singularity that occurs at z = 0 I use the following properties:

1) If $\stackrel{lim}{z \rightarrow z_{o}} |f(z)| \leq M$ (ie. remains bounded), $f(z)$ has a removable singularity at $z = z_{o}$.

2) If $\stackrel{lim}{z \rightarrow z_{o}} |f(z)| \rightarrow \infty$, $f(z)$ has a pole at $z = z_{o}$.

So I consider the limit of this function as $z \rightarrow z_{o}$.

$$\stackrel{lim}{z \rightarrow z_{o}} |f(z)| = \stackrel{lim}{z \rightarrow 0} |\frac{e^{z} - 1}{z}| = |\frac{e^{0} - 1}{0}| = |\frac{0}{0}| = \frac{0}{0}$$

So here I apply L'Hopital's Rule setting $f(z) = \frac{g(z)}{h(z)}$ with $g(z) = e^{z} -1$ and $h(z) = z$:

$$\stackrel{lim}{z \rightarrow z_{o}} |f(z)| = \stackrel{lim}{z \rightarrow z_{o}} |\frac{g(z)}{h(z)}| = \stackrel{lim}{z \rightarrow z_{o}} |\frac{g'(z)}{h'(z)}| = \stackrel{lim}{z \rightarrow z_{o}} |\frac{e^{z}}{1}| = |\frac{e^{0}}{1}| = |\frac{1}{1}| = |1| = 1$$

So we have a removable singularity of value 1 at $z = 0$. This result agrees with the final answer in the text, but is this approach valid? Below is my approach to a question with a pole (it is similar and so will be shorter).

"...locate each of the isolated singularities of the given function and tell whether it is a removable singularity, a pole, or an essential singularity (case (3)). If the singularity is removable, give the value of the function at the point; if the singularity is a pole, give the order of the pole.

5. $\frac{2z + 1}{z + 2}$" (Complex Variables, 2nd edition; by Stephen D. Fisher, pg. 150)

This function is undefined at $z = -2$ so we find the limit as z approaches this point:

$$\stackrel{lim}{z \rightarrow z_{o}} |f(z)| = \stackrel{lim}{z \rightarrow -2} |\frac{2z + 1}{z + 2}| = |\frac{-3}{0}| \rightarrow \infty$$

Therefore there is a pole at $z = -2$. The order of the pole for a function $f(z)$ is the order of zero at that point for a function $g(z) = \frac{1}{f(z)}$.

$g(z) = \frac{z+2}{2z -1}$ has a zero of order one at $z = -2$ and therefore $f(z)$ has a pole of order one at $z = -2$.

Is this an acceptable way to approach these problems? If yes, can I approach a function $f(z) = \frac{z^{2}}{sin z}$ in this way? I realize that if either of my above conditions (1 or 2) hold that there will be an infinite number of removable singularities or poles (and so I suspect it is more likely an essential singularity). Many thanks in advance!

2. Jul 7, 2013

### BruceW

your approach so far looks good to me. For your function z^2/sin(z) don't be too quick to jump to conclusions. Keep in mind that each of the singularities occurs at a different value of z.

3. Jul 7, 2013

### Tsunoyukami

I'll be sure to check that either condition one or two holds for "each" singularity. I suppose it would be possible that some of the singularity are removable while others are poles, correct? Thanks for taking a look! :)

4. Jul 7, 2013

### BruceW

have a look and find out :) that's what maths is all about, right? (well I wouldn't know, I learned physics really).

5. Jul 7, 2013

### Tsunoyukami

I'll work on that discovery when I get home from work! (and I'm a physics student too!) thanks again.

6. Jul 7, 2013

### BruceW

no problem! :)