Singularities and Analyticity at z=0

[Ted123]

Homework Statement

The Attempt at a Solution

Both $\displaystyle \frac{\cos(z)-1}{z^2}$ and $\displaystyle \frac{\sinh(z)}{z^2}$ have 1 singular point at $z=0$.

For (a):

z=0 is a removable singularity since defining f(0)=1 makes it analytic at all $z\in\mathbb{C}$.

z=0 is isolated since f(z) is analytic for 0<|z|<1. But z=0 is not a pole since cos(0)-1 =0, and so z=0 is an essential singularity.

For (b):

z=0 is a removable singularity since defining f(0)=1 makes it analytic at all $z\in\mathbb{C}$.

z=0 is isolated since f(z) is analytic for 0<|z|<1. But z=0 is not a pole since sinh(0)=0, and so z=0 is an essential singularity.

Is this correct?

 

[Dick]

Parts of it might be true. You said basically the same thing about both functions and you didn't prove anything you said. Give some arguments. If f(z)=sinh(z)/z^2, why does defining f(0)=1 make it analytic on C?

 

[Ted123]

Parts of it might be true. You said basically the same thing about both functions and you didn't prove anything you said. Give some arguments. If f(z)=sinh(z)/z^2, why does defining f(0)=1 make it analytic on C?

Probably because I'm not understanding the definitions correctly!

These are my set of definitions:

I think for both (a) and (b), z=0 is an isolated singularity but not a pole, so an essential singularity. But they probably aren't removable.

 

[Dick]

Probably because I'm not understanding the definitions correctly!

I think for both (a) and (b), z=0 is an isolated singularity but not a pole, so an essential singularity. But they probably aren't removable.

The definitions will be clearer to you if you look at a power series expansion of each function around z=0.

 

[Ted123]

The definitions will be clearer to you if you look at a power series expansion of each function around z=0.

I don't like how some of these definitions are given so if I use this definition of pole:

Clearly $z_0=0$ is an isolated singularity since it is the only singularity for both (a) and (b).

(a) $\displaystyle \lim_{z\to 0} \;(z-0)^N f(z) = \lim_{z\to 0} \; z^{N-2} (\cos(z)-1) = 0 \;\; \forall \;N>0$ so $z_0=0$ is not a pole. Hence it is an essential singularity.

(b) If N=1 then $\displaystyle \lim_{z\to 0} \;(z-0) f(z) = \lim_{z\to 0} \frac{\sinh(z)}{z} = 1 \neq 0$ so [itex]z_0=0[/tex] is a simple pole (of order 1). What would be the strength of the pole? It is not an essential singularity.

I'm not understanding how to see if 0 is a removable singularity in each case?

 

[Dick]

I don't like how some of these definitions are given so if I use this definition of pole:

Clearly $z_0=0$ is an isolated singularity since it is the only singularity for both (a) and (b).

(a) $\displaystyle \lim_{z\to 0} \;(z-0)^N f(z) = \lim_{z\to 0} \; z^{N-2} (\cos(z)-1) = 0 \;\; \forall \;N>0$ so $z_0=0$ is not a pole. Hence it is an essential singularity.

(b) If N=1 then $\displaystyle \lim_{z\to 0} \;(z-0) f(z) = \lim_{z\to 0} \frac{\sinh(z)}{z} = 1 \neq 0$ so [itex]z_0=0[/tex] is a simple pole (of order 1). What would be the strength of the pole? It is not an essential singularity.

I'm not understanding how to see if 0 is a removable singularity in each case?

You know how to expand cos(z) and sinh(z) in a power series around z=0. Put those expansions into the two functions and simplify. See what you think. Then look back at the definitions.