# Prove convergence of analytic part and principal part of laurent series

Let $$f(z)$$ be a holomorphic function in an annulus $$A(z_0,r_1,r_2)$$ with $$0\leq r_1\leq r_2$$
Consider the Laurent expansion $$f(z) = \sum_{n=-\infty}^{\infty} a_n (z-z_0)^n$$
How do I show that the analytic part and principal part of the series converges for $$|z-z_0| and $$|z-z_0|>r_1$$ respectively?

What I tried to do:
For Analytic part; $$\sum_{n=0}^{\infty} a_n (z-z_0)^n$$
Consider $$\Omega = D(0,r_2)$$
$$f(z)$$ is holomorphic in $$\Omega$$ , so we can rewrite f(z) as
$$f(z) = c_0 +c_1 (z-w) +c_2 (z-w)^2 + ... = \sum_{n=0}^{\infty} c_n (z-w)^n$$ where the series converges for $$|z-w|< r_2$$
Take $$w=z_0$$
Then $$\sum_{n=0}^{\infty} c_n (z-w)^n = \sum_{n=0}^{\infty} c_n (z-z_0)^n =\sum_{n=0}^{\infty} a_n (z-z_0)^n$$ by uniqueness of Laurent series
so we say that the Analytic part converges for $$|z-z_0|

For the principal part,
we do by the same argument?

Can somebody please tell me I'm on the right track? Or I should consider using $$limsup$$ etc to prove the convergence?

• What is the definition of Laurent expansion that you're using?
– user149874
Sep 27, 2015 at 6:44
• definition of laurent expansion? sorry I don't what u mean.... Sep 27, 2015 at 6:51
• I mean, what exactly do you mean when you say$f(z)=\sum a_n(z-z_0)^n$? Are the $a_n$ defined in terms of integrals, derivatives, or something else?
– user149874
Sep 27, 2015 at 6:58
• ah... $a_n = \frac{1}{2\pi i} \int_{C(z_0 ,r)} \frac{f(w)}{(w-z_0)^{n+1}} dw$ where $n = 0,+- 1,+-2,..., r_1 <r<r_2$ Sep 27, 2015 at 7:08

I'm going to take $z_0=0$ for simplicity. For $\rho_1, \rho_2$ with $r_1 < \rho_1 < \rho_2 < r_2$, and for $z$ with $\rho_1 < |z| < \rho_2$, Cauchy's integral formula applied to the annulus $\{ \rho_1 \leq |w| \leq \rho_2\}$ gives us that

$$f(z)= \frac{1}{2\pi i} \int_{|w|=\rho_2} \frac{f(w)}{w-z} \ \mathrm{d}w - \frac{1}{2 \pi i} \int_{|w|=\rho_1} \frac{f(w)}{w-z} \ \mathrm{d}w.$$

In the first integral we can write

$$\frac{f(w)}{w-z} = \sum_{n=0}^\infty \frac{z^nf(w)}{w^{n+1}},$$

and for each $z$ with $|z| < \rho_2$ this sum converges aboslutely uniformly on the circle $\{|w|=\rho_2\}$.

[Added in response to comment: there exists a positive real number $M$ such that $|f(w)| \leq M$ on $\{|w|=\rho_2\}$. Then for all $w, z$ with $|w|=\rho_2$ and $|z|<\rho_2$, and all $n \geq 0$, we have

$$\Bigg|\frac{z^nf(w)}{w^{n+1}}\Bigg| \leq \frac{M}{\rho_2} \Bigg(\frac{|z|}{\rho_2}\Bigg)^n.$$

The right-hand side converges (it's a geometric series with ratio $|z|/\rho_2 <1$) so the claimed sum converges absolutely uniformly on the circle by the Weierstrass M-test.]

So the first integral is equal to

$$\sum_{n=0}^\infty z^n \frac{1}{2\pi i} \int_{|w|=\rho_2} \frac{f(w)}{w^{n+1}} \ \mathrm{d}w,$$

and this converges and is valid on the whole of $\{|z| < \rho_2\}$. Letting $\rho_2 \rightarrow r_2$, we see that the positive part of the Laurent series converges on $\{|z| < r_2\}$.

For the negative part, do a similar expansion with $w^n/z^{n+1}$.

• sorry, but how do you know that for each $z$ with $|z|<\rho_2$ , the sum converges absolutely uniformly? Sep 27, 2015 at 7:55
• @marco I've added an explanation to my answer.
– user149874
Sep 27, 2015 at 8:19