# Thread: why this point is not essential singular point

1. ## why this point is not essential singular point

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

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
??

2. Originally Posted by transgalactic
$f(x)=\frac{e^{\frac{1}{z-1}}}{e^z-1}$
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

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
??
If $f(z)$ has a zero of order $m$ at $z_0$ then $\frac{1}{f(z)}$ has a pole of order $m$ at $z_0$. Now use this with $f(z)=e^z-1$. Note however that the function does have an essential singularity at $z=1$. Another thing, your argument is wrong since there is no $+\infty$ or $-\infty$ when dealing with complex functions (only a single point $\infty$).

3. thanks