1. ## Complex Singularities

Bonjour!

I have a few math questions involving locating and classifying singularities for the following functions:

a) $f(z) = \frac{1}{1-e^{z^{2}}}$

b) $f(z) = \frac{z}{e^{\frac{1}{z}}}$

c) $f(z) = \frac{1-e^{z}}{z}$

Merci!

2. Hint: Singularities occur when the denominator is $0 + 0i$.

3. Originally Posted by ComplexXavier
Bonjour!

I have a few math questions involving locating and classifying singularities for the following functions:

a) $f(z) = \frac{1}{1-e^{z^{2}}}$

b) $f(z) = \frac{z}{e^{\frac{1}{z}}}$

c) $f(z) = \frac{1-e^{z}}{z}$

Merci!
a) Poles of order 2 at values of z such that $1 - e^{z^2} = 0$.

b) Essential singularity at z = 0.

c) Removable singularity at z = 0.

I suggest you research the reasons why ....

4. I understand b and c now and how we came about finding them.

For a, I have found the singularities to be +/- sqrt(2*i*pi*n) but I am having trouble showing that these are poles of order 2. Any thoughts?

5. What reason do you have to think that they are poles of order 2?

6. Well, I tried to prove it to be removable or essentially and from my findings it was neither. So by process of elimination the only one I can't disprove is poles.

7. Originally Posted by ComplexXavier
I understand b and c now and how we came about finding them.

For a, I have found the singularities to be +/- sqrt(2*i*pi*n) but I am having trouble showing that these are poles of order 2. Any thoughts?
Can you show that z = 0 is a pole of order 2? There is a useful theorem in the second paragraph here: http://mathworld.wolfram.com/Pole.html (I'm sure the theorem is also in your class notes and textbook).