Let $\varphi$ be the Euler totient function, and let us define the function $f(z)$ by the series $$ f(z) := \sum_{n=1}^{\infty} \varphi(n) z^n $$ Since $0\le \varphi(n)\le n$, I believe this gives a well-defined function in some region of the complex plane. Where could I find a discussion on the analytic behavior of this function as a function of the complex variable $z$?

I am in particular interested in the pole structure as we approach $z=1$ from below ($z\to 1^{-}$), say on the real axis. More specifically let us write $z=e^{-\beta}$. Then for $\beta$ positive I would expect the function $f(\beta)$ to converge. Then I am interested in the limit $\beta\to 0^+$ from above, and especially in the constant part $c_0$: $$ f(z=e^{-\beta})=\frac{c_n}{\beta^n} + \cdots +\frac{c_1}{\beta} +c_0 + c_{-1} \beta +\cdots . $$

PS: as suggested by the comment, a potential clue is that there is a known identity $$ g(s):=\sum_{n=1}^{\infty} \frac{\varphi(n)}{n^s} =\frac{\zeta(s-1)}{\zeta(s)} $$

PS^2: As is commented in the answers, it indeed seems to be the case that $f(z)$ has the unit circle as natural boundary. However we can still ask the question of what the value of $c_0$ is, at least in certain cases. As an example consider a different function, which indeed takes the form of a Lambert series: $$ h(z):=\sum_{n=1}^{\infty} \varphi(n) \frac{z^n}{1-z^n} $$ This has unit circle as natural boundary. But then there is a nice result that this can be written as $$ h(z)=\frac{z}{(1-z)^2} $$ and hence $$ h(z=e^{-\beta})=\frac{1}{\beta^2}-\frac{1}{12}+\cdots $$ so we learn that $c_0=-\frac{1}{12}$ in this case. What I mean is that I would like to extract a similar number $c_0$ for the function $f(z)$. It indeed is not clear if this question has a well-defined answer for $f(z)$.