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Is the function $\frac{e^z-1}{e^z+1}$ analytic?

I tried using $\mathop{\lim}\limits_{z\to a} \frac{f(z)-f(a)}{z-a}$, but I didnt really know what to use for $a$.

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    This is $\tanh(z/2)$.2010-12-09
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    Dear @user4577: The numerator and denominator are both analytic, and the quotient of two analytic functions is analytic wherever the denominator is zero (by the same argument used in proving the quotient rule in the case of real functions). So this should help you.2010-12-09
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    Exponentials are analytic, constants are analytic, sums of analytic functions are analytic and so are quotients if the denominator is nowhere zero.2010-12-09
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    Remember that $\exp(\pi \imath) =-1$2010-12-09
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    The denominator is zero at an infinite, discrete set of points (overkill explanation: since the nonconstant entire function $\operatorname{exp}(z)$ misses the value $0$, it cannot also miss the value $-1$) at which the function is not defined. Away from these, it is analytic.2010-12-09
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    This is certainly homework somewhere :-) http://empslocal.ex.ac.uk/people/staff/rjchapma/courses/an10/an4.pdf2010-12-09
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    This function is a quotient of two holomorphic functions, so it is a meromorphic function. In particular, it is holomorphic (and therefore analytic) in every point in which it is defined.2011-03-16
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    Isn't it meromorphic??? Why are people saying it's analytic2011-04-15
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    @quanta: A meromorphic function is the same thing as a analytic function with codomain $\mathbb{C}_\infty$ (the Riemann sphere). Equally, it's an analytic function on the domain with all the poles removed.2011-05-15

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Yes, if $G$ is the set where the denominator is nonzero, then $f(z)=\frac{e^z-1}{e^z+1}$ is analytic on $G$. This is true because $e^z$ is analytic, constants are analytic, sums of analytic functions are analytic, and quotients of analytic functions are analytic away from zeros of the denominators (and the usual quotient rule for differentiation holds). The set where the denominator is $0$ is $\{(2n+1)\pi i:n\in\mathbb{Z}\}$, and $f$ has a simple pole at each point in this set.

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if $z=i\pi$ we have problems with denominator because $e^z+1=e^{i\pi}+1=-1+1$ and remember thqt the exponencial function is a periodic function, then $\frac{e^z-1}{e^z+1}$ is not analytic in the point where $e^z+1=0$