Consider the following question which was asked in my Complex analysis Quiz( which is now over) :
Consider the following analytic function $f(z)= \sum_{n=1}^{\infty}\frac{1+c}{n+c} z^n $ Determine the largest domain to which f can be anaytically continued?
Determine an analytic continuation of f from the unit disk to a larger domain?
I was not able to do any of the 2 questions and thought about it again today and not getting any idead except that the Radius of convergence of the series in |z|<1 but I think that this will not be a boundation in decieding analytic continuation as I have seen examples of series with ROC 1 but which can be continued to regions with |z|>1.
So, I request your help as I will not be able to do it myself. Thanks! I have been following Serge's Lang Complex Analysis.
There is no single answer. Consider e.g. $c=0$, where $\sum_{n=1}^\infty z^n/n = -\log(1-z)$ for $|z|<1$. This can be analytically continued to the complement of a branch cut, which can be any curve extending from $1$ out to $\infty$ and not intersecting the unit disk. But none of these is "the largest".
EDIT: In general, $f(z)$ can be written in terms of the Lerch Phi function. It satisfies the differential equation $$f'(z) + \frac{c}{z} f(z) = \frac{1+c}{1-z}$$ in the unit disk, i.e. $$ \dfrac{d}{dz} z^c f(z) = z^c \frac{1+c}{1-z}$$ Take any simply-connected open region $U$ of $\mathbb C$ not including $0$ or $1$, and an analytic antiderivative $F(z)$ of $z^c (1+c)/(1-z)$ on $U$, and $f(z) = z^{-c} (F(z) + constant)$ (for a suitable value of the constant) will be an analytic continuation of $f$ on that region.