Is z_0 a Removable Singularity in Complex Analysis?

[alexmahone]

By estimating the coefficients of the Laurent Series, prove that if $z_0$ is an isolated
singularity of $f$, and if $(z−z_0)f(z)\rightarrow 0$ as $z\rightarrow z_0$, then $z_0$ is removable.

My attempt

$\displaystyle f(z)=\sum\limits_{n=0}^\infty a_n(z-z_0)^n+\frac{b_1}{z-z_0}+\frac{b_2}{(z-z_0)^2}+\ldots+\frac{b_n}{(z-z_0)^n}+\ldots$

in a punctured disk $0<|z-z_0|<R_2$.

So, $\displaystyle(z-z_0)f(z)=\sum\limits_{n=0}^\infty a_n(z-z_0)^{n+1}+b_1+\frac{b_2}{z-z_0}+\ldots+\frac{b_n}{(z-z_0)^{n-1}}+\ldots$

in $0<|z-z_0|<R_2$.

How do I proceed?

 

[jvicens]

To prove that $z_0$ is a removable singularity, we need to show that the Laurent series of $f(z)$ around $z_0$ has no negative powers of $(z-z_0)$. This means that all the coefficients $b_n$ must be equal to 0.

Since $(z-z_0)f(z)\rightarrow 0$ as $z\rightarrow z_0$, we know that the Laurent series of $f(z)$ must converge to 0 at $z=z_0$. This implies that $\lim_{n\rightarrow\infty}a_n(z-z_0)^n=0$ as $z\rightarrow z_0$.

Using the fact that $a_n$ is the coefficient of the $(z-z_0)^n$ term in the Laurent series of $f(z)$, we can write this as $\lim_{n\rightarrow\infty}a_n=0$. This means that all the coefficients of the positive powers of $(z-z_0)$ in the Laurent series of $f(z)$ must be equal to 0.

Since all the positive powers of $(z-z_0)$ are eliminated, the Laurent series of $f(z)$ reduces to a polynomial in $(z-z_0)$, which is analytic at $z=z_0$. This means that $z_0$ is a removable singularity of $f(z)$.