Singularities of $f:\mathcal{D}^*\rightarrow \mathbb{C}$ $\mathcal{D}^*=\{z: 0 < \mid z \mid < 1 \}$ $\mathcal{D}^*=\{z: 0 < \mid z \mid < 1 \}$ and $f:\mathcal{D}^*\rightarrow \mathbb{C}$ such that Re$f(z)>-10$ for all $z$.
Could someone please explain why:
(i) By Weierstrass-Casorati theorem, $f$ cannot have an essential singularity at $0$
(ii) $f$ has no pole at $0$
 A: Hint for (ii): Suppose $f$ has a pole of order $n>0$ at $0.$ Then for $z\in D^*,$
$$f(z) = \frac{a_n}{z^n} + \cdots + \frac{a_1}{z} + g(z),$$
where $a_n\ne 0$ and $g$ is analytic in $D(0,1).$ If you choose $t_0$ correctly, then $a_n/(re^{it_0})^n$ is negative for $r\in (0,1)$ and $\to - \infty.$ This term dominates in the behavior of $f(re^{it_0})$ as $r\to 0^+.$
A: First of all: Since $f$ has a pole / essential singularity at zero if and only if $f + 10$ has, we may instead assume $\operatorname{Re} f > 0$ and, in particular, that $f$ has no zeroes on $\mathcal D^\ast$.
(i) If $f$ had an essential singularity at zero, then $f(\mathcal D^\ast)$ would be dense in $ℂ$ by Casaroti–Weierstraß, but $\operatorname{Re} f(\mathcal D^\ast)$ is always positive and so all points with negative real part are isolated from $f(\mathcal D^\ast)$.
(ii) If $f$ had a pole at zero, $1/f$ could be extended to all $\mathcal D$, vanishing at zero. But as $\operatorname{Re} (1/f)$ would then be an open map on $\mathcal D$, that would contradict $\operatorname{Re} (1/f) > 0$ on $\mathcal D^\ast$ (which comes from $\operatorname{Re} f > 0$).
You can also do both simultaneously, without using Casorati–Weierstraß, only using Riemann’s theorem on removable singularities.
There is a Möbius transformation $ρ \colon \mathcal H → \mathcal D,~z ↦ \frac{z - \mathrm i}{-\mathrm iz + 1}$, which biholomorphy maps the upper half plane $\mathcal H = \{z ∈ ℂ;~\operatorname{Im z} > 0\}$ onto the unit disk $\mathcal D$. Since $\operatorname{Re} f > 0$, its rotation $\mathrm i f$ maps into $\mathcal H$.
Therefore, $ρ∘(\mathrm if)$ is bounded on $\mathcal D^\ast$, so its singularity at zero may be removed and $ρ∘(\mathrm if)$ can be extended to a holomorphic map $\mathcal D → \mathcal D$. By multiplying with $ρ^{-1}$ and dividing by $\mathrm i$, $f$ too can be extended to a holomorphic map on $\mathcal D$, so the singularity at zero is removable.
