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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 $|z-z_0| <r_2$ 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 $|z-z_0| <r_2$

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?

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  • $\begingroup$ What is the definition of Laurent expansion that you're using? $\endgroup$
    – user149874
    Sep 27, 2015 at 6:44
  • $\begingroup$ definition of laurent expansion? sorry I don't what u mean.... $\endgroup$
    – marco
    Sep 27, 2015 at 6:51
  • $\begingroup$ 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? $\endgroup$
    – user149874
    Sep 27, 2015 at 6:58
  • $\begingroup$ 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 $ $\endgroup$
    – marco
    Sep 27, 2015 at 7:08

1 Answer 1

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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 $w^n/z^{n+1}$.

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

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