# Show that function has removable singularity

Let $a\in\Omega$ be an isolated singularity of function $f$ that is holomorphic on $\Omega \setminus \{a\}$. Show that if $\mathbb{Re}f(z)>0$ in some neighbourhood of point $a$, then $a$ is a removable singularity for $f$.

$f$ has removable singularity at $a$ if and only if $\lim_{z\rightarrow a}(z-a)f(z)=0$.

But I don't know how to start. Do you have any idea?

• Consider the biholomorphic map $\phi$ from the half plane $\Re w>0$ to the unit disc $\lvert\zeta\rvert<1$. Study the composite $\phi\circ f$. – Yai0Phah Jan 16 '15 at 12:20
• Sorry, but I dont see how does it help... – luka5z Jan 16 '15 at 12:24
• Let $F=\phi\circ f$, then what about $\lim_{z\to a}(z-a)F(z)$? What can you conclude? Then use $f=\phi^{-1}\circ F$ to obtain what you want. – Yai0Phah Jan 16 '15 at 12:31
• $\lim_{z\to a}(z-a)F(z) = 0$? – luka5z Jan 16 '15 at 12:40
• You can say that it is not an essential singularity for sure, because of the Picard theorem: en.wikipedia.org/wiki/Picard_theorem – Igor Jan 16 '15 at 19:27

## 1 Answer

The function $$g(z) = \frac{f(z) - 1}{f(z) + 1}$$ has an isolated singularity at $a$ and satisfies $|g(z)| < 1$ in a (deleted) neighborhood of $a$, thus $\lim_{z\to a} g(z)$ exists. Hence $$\lim_{z\to a} (z - a)f(z) = \lim_{z\to a} (z - a)\frac{1 + g(z)}{1 - g(z)} = \lim_{z\to a} (z - a)(1 + g(z) + O(g(z)^2)) = 0.$$ Consequently, $a$ is a removable singularity of $f$.

• Is it not possible that the limit of $g(z)$ is 1? – zokomoko Feb 3 '17 at 16:51
• Yes, it's possible. – kobe Feb 3 '17 at 19:20
• Thank you for your reply. Could you please elaborate on how this limit turns out to be zero? The denominator (if $g$ approaches 1) and the numerator both vanish.. – zokomoko Feb 3 '17 at 21:31
• Look at the expansion $(z - a)(1 + g(z) + O(g(z)^2))$. We know $(z - a)(1 + g(z)) \to 0$ as $z \to a$, and since $\lvert g(z)\rvert < 1$ in a deleted neigborhood of $a$, then $(z - a)O(g(z)^2) \to 0$ as $z \to a$. – kobe Feb 3 '17 at 21:33