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 $a$.
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 $a$.
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.
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 $e^z+1=0$