2.5 Q19

[Huanglei Ln]

Can anyone help me this question? Thanks!
Suppose that the Laurent series $\sum\nolimits_{-\infty}^{+\infty}a _n (z-z_0)^n $ converges for  $ f<| z-z_0|<R $ and
$$
\sum \limits_{-\infty}^{+\infty}  a_n (z-z_0)^n=0, 0<|z-z_0|<r.
$$

Show that $a_n=0,n=0, \pm 1, \pm 2,\cdots$, (Hint:Multiply the series by $(z-z_0)^{-m}$ and integrate around the circle $|z-z_0|=s$, $r<s<R$ with respect to $z$. The result must be zero, but it is also $a_{m-1}$.)

[Victor Ivrii]

Well, there was a hint provided:if
$$
f(z)=\sum _{n=-\infty}^\infty a_n (z-z_0)^n =0
$$
as $|z-z_0|=s=$ with $s\in (r,R)$ then
$$
0=\int_\gamma f(z)(z-z_0)^{-m-1}\,dz =\sum _{n=-\infty}^\infty \int_\gamma a_n (z-z_0)^{n -m-1}\,dz
$$
while $\int_\gamma  (z-z_0)^{k}\,dz=0$ for $k\ne -1$ and $2\pi i$ for $k=1$.

[Yunfei Xia]

Follow the hint: let $r$ be the circle $|z-z_0|=s,0<s<r$. For any $m\in z$.
\begin{align*}
0&=\frac{1}{2\pi i} \int_r(z-z_0)^{-m}\left(\sum\limits_{-\infty}^{+\infty}a_n(z-z_0)^n\right)\text{d}z\\
&\quad z=z_0+se^{i\theta},\quad
z-z_0=se^{i\theta}\\
&\quad \text{d}z=sie^{i\theta}\text{d}\theta\\
&=\frac{1}{2\pi }\int\nolimits_{0}^{+2 \pi}\left(se^{i\theta}\right)^{-m}\left(\sum\limits_{-\infty}^{+\infty}{a_n}\left(se^{i\theta}\right)^n\right)se^{i\theta}\text{d}\theta\\
&=\sum\limits_{-\infty}^{+\infty}{a_n}s^{-m}s^{n+1}\frac{1}{2\pi }\int e^{-im \theta+i \theta+in}\text{d}\theta\\
&=\sum\limits_{-\infty}^{+\infty}{a_n}s^{-m}s^{n+1}\frac{1}{2\pi} \int e^{i \theta(n+1-m)}\text{d}\theta\\
&\text{Let}~n+1-m=0,n=m-1\\
&=a_{m-1}.
\end{align*}

[Victor Ivrii]

not clear what do you mean by "let $n=...$". $n$ runs from $-\infty$ to $\infty$ and you must check all values

[Ende Jin]

What is the first condition used for? The one that the series converges in the $\{z: r < |z| < R\}$.

[Victor Ivrii]

If we don't know that series is converging on the circle, we integrate along, the integration is senseless

[Ende Jin]

I realized that you provided a solution different from the given one back in the book.
But the question doesn't say $\sum a_n(z-z_0)^n = 0$ when $r < |z-z_0| < R$. It just converges. The series converges to zero only if $0 < |z - z_0| < r$.
That is basically my question. The first condition doesn't seem helpful.

[Victor Ivrii]

Ende Jin
Please provide exact citation from the textbook.

[Ende Jin]

It is at P406, The solution for Q19.
Actually, Yunfei Xia provided a very similar proof with that.