# Residue of complex function $f$

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 $$\lim a_{n}$$.

• Hint: $f(z)=\frac c{z-1}+g(z)$, where $g$ has a removable singularity at $1$. 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

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

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

• clever idea... hmph Jun 21 at 11:20