show that there exits a analytic cube root of a function on a domain Show that $f(z)=e^z+\frac{1}{(z-1)^3}+\frac{1}{(z+1)^6}$ has an analytic cube root on $\mathbb C$-{1,-1}.
The usual way is to pick up $ z_0 \notin \mathbb C$-{1,-1}, and $\forall z \in \mathbb C$-{1,-1}, let $\gamma_{z_0}^z$ be any curve that does not go through {1,-1}, then we can define $g(z)=e^{\frac{1}{3}\int_{\gamma _{z_0}^z}\frac{f'}{f}}$. To show g is analytic is to show g is well defined, or in other words, show $\frac{1}{3}\int_{\gamma }\frac{f'}{f}$ is a multiple of 2$\pi i, \space$ for any closed curve $\gamma$ with endpoint $z_0$ . The problem here is that although the number of poles (considering the multiplicity) of $f$ is a multiple of three, but f has two zeros in $\mathbb C$-{1,-1}. So by argument principle, I cannot see  $\frac{1}{3}\int_{\gamma }\frac{f'}{f}$ is a multiple of 2$\pi i, \quad$ for any closed curve $\gamma$ with endpoint $z_0$ 
Can anyone give me some ideas? Thanks in advance.
 A: If a meromorphic function $f \not\equiv 0$ defined on a domain $U \subset \mathbb{C}$ has a meromorphic $k$-th root, then the multiplicity of all zeros and poles of $f$ must be a multiple of $k$ (if $U$ is simply connected, that is also sufficient).
The given function
$$f(z) = e^z + \frac{1}{(z-1)^3} + \frac{1}{(z+1)^6}$$
is real ($f(\mathbb{R}) \subset \mathbb{R}\cup \{\infty\}$), and its only two poles are at $-1$ and $1$. For real $x < 1$, the summand $\frac{1}{(x-1)^3}$ is negative, and the summands $\frac{1}{(z+1)^6}$ and $e^z$ are positive. Since the negative summand vanishes of lower order than the positive summands, $f(x)$ is negative for negative $x$ of large absolute value. Since the pole at $-1$ is of even order, $f(x)$ is positive for real $x$ close to $-1$, and $f(x) < 0$ for $x$ close to $1$ but smaller than $1$. Thus $f$ has a zero $x_1$ in the interval $(-1,1)$ [at $\approx 0.145$], and a zero $x_2$ in the ray $(-\infty,-1)$ [at $\approx -5.760$]. It is easy to see that $f'(x) < 0$ for $-1 < x < 1$, and one can check that $f'(x_2) > 0$.
So the two known zeros [$f$ might have further non-real zeros] are simple, and $f$ cannot have an analytic cube root on any domain containing either of them. In particular, $f$ cannot have an analytic cube root on $\mathbb{C}\setminus \{1,-1\}$ nor on $\mathbb{C}\setminus [-1,1]$. If $f$ has no further zeros, or all further zeros have multiplicity divisible by $3$, then $f$ has an analytic cube root on $\mathbb{C}\setminus (-\infty,1]$, however.
