1
$\begingroup$

Let $D=\{z\in \mathbb{C} :|z|<2 \}$ and $f$ be a function on $D$ which is analytic at every point except $z=1$ and also it has a simple pole at $z=1$. Suppose that $$f(z)=\sum\limits_{n=0}^{\infty} a_{n}z^{n} , \qquad |z|<1.$$ Show that $\lim\limits_{n\to\infty} a_{n}=-c$ where $c$ is residue of $f$ at $z=1$.

My approach is that, define a Laurent expansion of f around $z=1$ and compare Taylor expansion and Laurent expansion at $z=1/2$ and compute $\lim a_{n}$.

$\endgroup$
1
  • 2
    $\begingroup$ Hint: $f(z)=\frac c{z-1}+g(z)$, where $g$ has a removable singularity at $1$. $\endgroup$ Jun 21 at 10:58
2
$\begingroup$

A simpler method uses the fact that reducing the residue of a simple pole to zero removes the singularity entirely.

To implement this, render

$f(z)=g(z)+\dfrac{c}{z-1}$

where the residue of $g$ at $1$ is reduced to zero leaving $g$ analytic there. We then have the following Taylor series expansions:

$f(z)=\sum_{n=0}^\infty a_nz^n$

$g(z)=\sum_{n=0}^\infty b_nz^n$

$\dfrac{c}{z-1}=-\sum_{n=0}^\infty cz^n$

$\color{blue}{a_n=b_n-c}$

Then $g$ is analytic through all of $|z|\le1$ forcing the Taylor series for $g$ to converge at $z=1$, $\therefore b_n\to0$ as $n\to\infty$. It follows that $a_n=b_n-c\to-c$.

$\endgroup$
1
  • 1
    $\begingroup$ clever idea... hmph $\endgroup$ Jun 21 at 11:20

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.