If $f$ has an isolated singularity at $z_0$ and if $\lim_{z \to z_0}(z-z_0)f(z) = 0$, then the singularity is removable. This is the Riemann Principle of removable singularity and I am using the proof of BaK and Newman. Below is one part which i do not understand.

\begin{equation*} h(z) = \begin{cases} (z-z_0)f(z) ~~~ z \neq z_0 \\ 0 ~~~~~~~~~~~~~~~~~~ z = z_0 \end{cases} \end{equation*}

Now by our hypothesis, we have $h$ to be continuous at the very point $z = z_0$. By how we define $h$, it is easy to see that $h$ is analytic at $z = z_0$, since there surely exists an $\epsilon$ neighborhood around $z_0$ such that $h$ is analytic since $f$ is analytic at $z \neq z_0$.


Since $h(z_0) = 0$, $g(z) = \dfrac{h(z)}{(z-z_0)}$ is analytic at $z_0$; I could not undersand why $g$ is analytic at $z_0$, isnt the function undefined at the denominator?

** and equals $f$ for $z \neq z_0$. Thus concluding the proof.


Since $h$ is analytic in a neighborhood of $z_0$ and $h(z_0)=0$ the power series expansion around $z_0$ is of the type $0+a_1(z-z_0)+a_2(z-z_0)^{2}+\cdots$ It is understood that in the suggested proof $g(z)$ is defined as $a_1+a_2(z-z_0)+a_3(z-z_0)^{2}+\cdots$ so that $g(z)=\frac {h(z)} {z-z_0}$ near $z_0$ (excluding the point $z=z_0$). Since $g$ is defined by a convergent power series it is analytic in a neighborhood of $z_0$.


Not a direct answer, but may help you intuitively.

Warm-up exercise

If $$g(z)=\sum^\infty_{-\infty}c_n (z-a)^n$$ and $$\lim_{z\to a}g(z)=0$$, then $$a_0=a_{-1}=a_{-2}=\cdots=0$$

Suppose the Laurent series of $f$ about $z_0$ is

$$\sum_{n=-\infty}^{\infty}a_n (z-z_0)^n$$


$$(z-z_0)f(z)=\sum^\infty_{n=-\infty}a_n (z-z_0)^{n+1}=\sum^\infty_{-\infty}a_{n-1}(z-z_0)^n$$

For $\lim_{z\to z_0} (z-z_0)f(z)=0$, as seen in the warm-up exercise, we have $a_{-1}=a_{-2}=a_{-3}=\cdots=0$.

Thus, indeed $f$ has a Taylor series at $z_0$, thus $f$ is analytic there, and the singularity there is called removable singularity.


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