2
$\begingroup$

I have the function: $$ f(z) = \frac{z^2 + 1}{z(1-z)}$$

Which I have to show that it is analytic on $\mathbb{C}\setminus \{0,1\}$ (from definition) so I must write it in the form: $$\sum_n a_n (z-z_0)^n$$

I can't use the fact that it is holomorphic because we haven't taken that rule yet.

I have managed to write $f$ as: $$f(z) = 1 - \frac{1}{z} + 2\frac{1}{1-z}$$

Then $$\frac{1}{z} = \frac{1}{z_0\left(1 + \frac{z-z_0}{z_0}\right)} = \frac{1}{z_0}\sum_n \left(\frac{z-z_0}{z_0}\right)^n$$ using the property of geometric series, and: $$\frac{1}{1-z} = \sum_n z^n$$ also using the geometric series property.

So I end up with: $$f(z) = 1 - \frac{1}{z_0}\sum_n \left(\frac{z-z_0}{z_0}\right)^n + 2\sum_n z^n$$

That's the closest I got to an answer.

Any help on this question is much appreciated

  • 0
    You can show that $f=\frac{g}{h}$ where $g, h$ are holomorphic functions is holomorphic in $\Bbb C - h^{-1}({0})$. – 2017-02-18
  • 0
    thank you but I cant use the rule that every holomorphic function is analytic, I need to prove it using definition – 2017-02-18

1 Answers 1

2

In your partial fraction decomposition you've made a mistake, actually

$$f(z) = -1 + \frac{1}{z} + \frac{2}{1-z}.$$

That doesn't touch the essence of the matter, however.

You started well, but then you expanded $\frac{1}{1-z}$ about the centre $0$ instead of $z_0$. Just write

$$\frac{1}{1-z} = \frac{1}{(1 - z_0) - (z-z_0)} = \frac{1}{1-z_0}\cdot \frac{1}{1-\frac{z-z_0}{1-z_0}}$$

and expand the latter in a geometric series (that converges for $\lvert z-z_0\rvert < \lvert 1-z_0\rvert$).

  • 0
    ah that's awesome, thank you. But there's no problem in keeping the -1 out? – 2017-02-18
  • 1
    A constant has a trivial power series expansion. At the end, you collect the three series into one and obtain $$f(z) = \sum_{n = 0}^{\infty} \biggl(\frac{1}{z_0^{n+1}} + \frac{2}{(1-z_0)^{n+1}} - \delta_{n,0}\biggr)(z-z_0)^n,$$ where $\delta$ is the Kronecker-$\delta$. You also need to state (and give an argument) that the series converges to $f$ on some disk about $z_0$. The series converges (to $f$) for $\lvert z-z_0\rvert < \min \{ \lvert z_0\rvert, \lvert 1-z_0\rvert\}$. – 2017-02-18
  • 0
    it's clear, thank you – 2017-02-18