# Why this point is not essential singular point

1. Jul 12, 2010

### nhrock3

$$f(x)=\frac{e^{\frac{1}{z-1}}}{e^z-1}$$
if we go $$z=0^+$$ f(z)=+infinity
if we go $$z=0^-$$ f(z)=-infinity
we dont have a limit here
so why its a pole
??

by my definition a pole is when $$f(z)=\frac{g(z)}{(a-z)^n}$$

if g is analitical at g(a) then 'a' in th n'th order pole

but in my case there is no (a-z)^n representation
and i proved that there is no limit

Last edited: Jul 12, 2010
2. Jul 12, 2010

### nhrock3

in my exam i have much complicated functions
and i cant develop it into a series every time

i tried to solve this one strictly by the limit tests
where am i wrong in my limit test
where is my conclusion wrong
??

3. Jul 12, 2010

### nhrock3

i was told that there is no +infinity and -infinity in complex functions
only infinity

but i have a solved question from a test which is solved exactly like it you said was wrong
and got full credit

maybe i am missing something here

4. Jul 12, 2010

### tmccullough

Functions don't have a "limit" at a simple pole (like this one), they have aribitarily large norm.

At poles, the values "wrap around" the complex plane at infinity, in other words, the function hits all complex numbers with arbitarily large norm. You always see a complex function going to the whole set of numbers $re^{i\theta}$ for $\theta\in [0,2\pi)$ and $r$ aribtraily large.

5. Jul 12, 2010

### nhrock3

i tried to develop it into a series

if my development is correct how whould you continue to develop it
so we will get a series with only one z variable

??
i cant make a guess here its all has to be series with only one z in a certain power in order to make a conclution

6. Jul 12, 2010

### nhrock3

i understand why
[MATH]\sin \left( \frac{1}{z} \right)[/MATH] has an essential singularity at [MATH]0[/MATH]

but how we can take any function we want and say sin(f) has significant point
at the pole of f
??
like here
[MATH]\sin ( f(z))[/MATH]
[MATH]f(z)= \frac{z}{z+1}[/MATH]