Prove function is complex analytic I want to prove that $f(z)=\frac{z}{e^z-1}$ is analytic around the origin. I tried using $z=x+iy$ and attempted to express $f$ as $u(x,y) + iv(x,y)$ to apply the Cauchy Riemann equations, but this is proving to be very cumbersome. Is there a better way to prove this?
 A: The Cauchy-Riemann equations are usually way more cumbersome to prove that a function is holomorphic than other methods. The usually most convenient way to see a function is holomorphic uses that sums, products, quotients (where the denominator is not 0, or more precisely where the denominator does not vanish with higher order than the numerator) and compositions of holomorphic functions are holomorphic, and that the sum of a convergent power series is holomorphic in the disk of convergence. Also, parameter-dependent integrals where the integrand depends holomorphically on the parameter are holomorphic when differentiation under the integral is legitimate.
Here, we can expand $e^z-1$ in a power series:
$$e^z-1 = \left(\sum_{n=0}^\infty \frac{z^n}{n!}\right)-1 = \sum_{n=1}^\infty \frac{z^n}{n!},$$
and that can evidently be divided by $z$ without problems, so
$$h(z) := \frac{e^z-1}{z} = \sum_{n=1}^\infty \frac{z^{n-1}}{n!} = \sum_{k=0}^\infty \frac{z^k}{(k+1)!}$$
is recognised as an entire holomorphic function, with $h(0) = 1 \neq 0$. Thus
$$\frac{z}{e^z-1} = \frac{1}{h(z)}$$
is holomorphic in a neighbourhood of $0$.
