# why this point is not essential singular point

• July 12th 2010, 09:37 AM
transgalactic
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
??
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• July 12th 2010, 09:58 AM
Jose27
Quote:

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
??
http://www.physicsforums.com/Prime/buttons/report.gif http://www.physicsforums.com/Nexus/misc/progress.gif

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 $\infty$).
• July 12th 2010, 10:08 AM
transgalactic
thanks :)