# At $z = 0$ the function $f(z)=(e^z+1)/(e^z-1)$ has what type of singularity

At $z = 0$ the function $f(z)=(e^z+1)/(e^z-1)$ has
1. A removable singularity
2. A pole
3. An essential singularity
4. The residue of $f (z )$ at $z = 0$ is $2$.

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do you know the definition of the concepts you mention in the options? – Ittay Weiss Dec 21 '12 at 4:51

Since the numerator is nonzero at $z=0$ but the denominator is $0$, we see that the singularity is not removable. If we write out the power series for $e^z$, we see that $$f(z)=\frac{e^z+1}{z+z^2/2+z^3/6+\cdots}=\frac{1}{z}\frac{e^z+1}{1+z/2+z^2/6+\cdots}$$ and since power series with nonzero constant term are invertible, this means that $zf(z)$ is analytic at $0$. Hence $f(z)$ has a simple pole at $z=0$. This also means the residue can be calculated as $1/(1/f(z))'|_{z=0}$, which is $$\left(\left.\left(\frac{e^z-1}{e^z+1}\right)'\right|_{z=0}\right)^{-1}=\left(\left.\frac{e^z(e^z-1)-e^z(e^z-1)}{(e^z+1)^2}\right|_{z=0}\right)^{-1}=\left(\frac{2}{4}\right)^{-1}=2$$

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1) Does the numerator $e^z+1$ have a root at $z=0$? if so, of what order? 2) Does the denominator $e^z-1$ have a root at $z=0$? if so, of what order? 3) What can you conclude about the quotient?

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0 is a pole of f as lim f at 0 does not exist but lim 1/f exist

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$1)$ $p(z)=e^z+1$ & $q(z)=e^z-1$ are both analytic at $z=0;$

$2)$ $p(0)\neq 0, q(0)=0, q^{(1)}(0)\neq0$ $\implies q$ has a zero at $0$ of order 1.

So $p(z)\over q(z)$ has a pole at $0$ of order 1.

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