removable singularity f(z) is analytic on the punctured disc $D(0,1) - {0}$ and the real part of f is positive. Prove that f has a removable singularity at $0$.

Instead of looking at $e^{-f(z)}$, I think it's easier to do the following.
First, assume that $f$ is non-constant (otherwise the problem is trivial).
Let $\phi$ be a conformal mapping (you can write down an explict formula for $\phi$ if you want) from  the right half-space  onto the unit disc, and let $g(z) = \phi(f(z))$. Then $g$ maps the punctured disc into the unit disc, so in particular $g$ is bounded near $0$, which implies that $g$ must have a removable singularity at $z = 0$. Also, by the open mapping theorem $|g(0)| < 1$.
On the other hand,
$$f(z) = \phi^{-1}(g(z))$$
and since $|g(0)| < 1$ and, and $\phi$ is continuous on the unit disc, the limit $\lim_{z\to 0} f(z) = \phi^{-1}(g(0))$ also exists, which means that $0$ is a removable singularity for $f$.
