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 $\lim a_{n}$.

  • 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

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

To implement this, render


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

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

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

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


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 $a_n=b_n-c\to-c$.

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

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