Why this point is not essential singular point

[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 don't 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

 

[nhrock3]

in my exam i have much complicated functions
and i can't 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
??

 

[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

 

[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.

 

[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 can't make a guess here its all has to be series with only one z in a certain power in order to make a conclution

 

[nhrock3]

$$\sin (z)$$ has an essential singularity at infinity, which means $$\sin \left( \frac{1}{z} \right)$$ has an essential singularity at $$0$$ (this is easy to check since the principal part of the Laurent expansion has infinitely many terms) and this gives that $$\sin ( f(z))$$ has an essential sing. wherever $$f$$ has a pole. Take $$f(z)= \frac{z}{z+1}$$ and the conclusion follows.

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

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