Is there a function analytic in $D(0,r)$ such that $(f(z))^2 = e^z + z$ for all $z\in D(0,r)$?

Question

I'm stuck on the following problem:

Let $D = \{z\in\mathbb{C}:|z|<r\}$, $r>0$. Is there a function analytic in $D$ such that $(f(z))^2 = e^z + z$ for all $z\in D$? Look at the two cases:

1. $r = 1$
2. $r = \frac{1}{2}$

I first wrote that for such a function, $$f(z) = (e^z + z)^{\frac{1}{2}} = e^{\frac{1}{2}\log(e^z + z)}$$ Then $f$ is analytic where $\log(e^z + z)$ is analytic. So my guess at the answer is that for the second case ($r=\frac{1}{2}$), $f$ is analytic in this disk because I can make a branch cut and make $\log(e^z + z)$ a single-valued function; and for the first case ($r=1$), there may be no such branch I can take.

However, I'm unsure of how to show this last part rigorously: I did some playing around with $e^z+z$ and am guessing that for the second case ($r=\frac{1}{2}$), I may want to try to show that I can take the principal branch because (maybe) $e^z+z$ cannot attain negative values in this disk. But I'm not sure where to go from here, and not sure what to do for the first case ($r=1$). Thanks for any answers!