0
$\begingroup$

Let $f(z)$ be a holomorphic function in an annulus $A(z_0,r_1,r_2)$ with $0<=r_1<=r_2$
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?

$\endgroup$
  • $\begingroup$ What is the definition of Laurent expansion that you're using? $\endgroup$ – Jez Sep 27 '15 at 6:44
  • $\begingroup$ definition of laurent expansion? sorry I don't what u mean.... $\endgroup$ – marco Sep 27 '15 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? $\endgroup$ – Jez Sep 27 '15 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 '15 at 7:08
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ sorry, but how do you know that for each $z$ with $|z|<\rho_2$ , the sum converges absolutely uniformly? $\endgroup$ – marco Sep 27 '15 at 7:55
  • $\begingroup$ @marco I've added an explanation to my answer. $\endgroup$ – Jez Sep 27 '15 at 8:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.