Toronto Math Forum
MAT3342018F => MAT334Lectures & Home Assignments => Topic started by: Huanglei Ln on November 18, 2018, 04:01:10 AM

Can anyone help me this question? Thanks!
Suppose that the Laurent series $\sum\nolimits_{\infty}^{+\infty}a _n (zz_0)^n $ converges for $ f< zz_0<R $ and
$$
\sum \limits_{\infty}^{+\infty} a_n (zz_0)^n=0, 0<zz_0<r.
$$
Show that $a_n=0,n=0, \pm 1, \pm 2,\cdots$, (Hint:Multiply the series by $(zz_0)^{m}$ and integrate around the circle $zz_0=s$, $r<s<R$ with respect to $z$. The result must be zero, but it is also $a_{m1}$.)

Well, there was a hint provided:if
$$
f(z)=\sum _{n=\infty}^\infty a_n (zz_0)^n =0
$$
as $zz_0=s=$ with $s\in (r,R)$ then
$$
0=\int_\gamma f(z)(zz_0)^{m1}\,dz =\sum _{n=\infty}^\infty \int_\gamma a_n (zz_0)^{n m1}\,dz
$$
while $\int_\gamma (zz_0)^{k}\,dz=0$ for $k\ne 1$ and $2\pi i$ for $k=1$.

Follow the hint: let $r$ be the circle $zz_0=s,0<s<r$. For any $m\in z$.
\begin{align*}
0&=\frac{1}{2\pi i} \int_r(zz_0)^{m}\left(\sum\limits_{\infty}^{+\infty}a_n(zz_0)^n\right)\text{d}z\\
&\quad z=z_0+se^{i\theta},\quad
zz_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+1m)}\text{d}\theta\\
&\text{Let}~n+1m=0,n=m1\\
&=a_{m1}.
\end{align*}

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

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

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

I realized that you provided a solution different from the given one back in the book.
But the question doesn't say $\sum a_n(zz_0)^n = 0$ when $r < zz_0 < R$. 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.

Ende Jin
Please provide exact citation from the textbook.

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