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.

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 $$z_0$$.

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

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

Then,

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

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.