1
$\begingroup$

We are given the function $f: \mathbb C \to \mathbb C$ defined by $f(z)=\frac{z-1}{z^2-3z+3}$

Is it possible to define $f$ as its taylor expansion near the point $z=i\sqrt 3$? If so, what is the radius of convergence? And find the taylor expansion of $f$ at that point.

I'm having some difficulties solving this question.

$z-1$ is analytic and so is $z^2-3z+3$ so $f$ should be analytic in any open region that does not contain any roots of $z^2-3z+3$. luckily for us, $i\sqrt 3$ is not a root, so there is an open disk with $i\sqrt 3$ as its center, where $f$ is analytic everywhere in that disk.

It can be also shown that the radius of this disk is $\sqrt 3$.

My only problem is actually finding the taylor expansion of $f$. I can't seem to find a nice form for the derivative.

$f'(z)=\frac{ -z^2+2z}{(z^2-3z+3)^2}$ and the second derivative looks even worse. Is there a more clever way of doing this?

$\endgroup$
4
  • $\begingroup$ Have you tried working with the variable $\zeta=z+i\sqrt{3}$ around $\zeta=0$? Note that since your function is of the form $P(z)/Q(z)^2$ with $P$ and $Q$ polynomials, the task somehow reduces to finding the Taylor expansion of $1/Q(z)$. $\endgroup$ Jan 8, 2015 at 9:21
  • $\begingroup$ I understand that we only need to find the taylor of $\frac{1}{Q(z)}$, but I lost you at the variable change. $\endgroup$ Jan 8, 2015 at 9:29
  • 1
    $\begingroup$ I suppose a typo in Andrea Mori's comment.$\zeta=z-i\sqrt{3}$ is a good suggestion. $\endgroup$ Jan 8, 2015 at 9:49
  • $\begingroup$ Ah, yes ... I meant to write $z=\zeta+i\sqrt{3}$. $\endgroup$ Jan 8, 2015 at 10:57

1 Answer 1

1
$\begingroup$

Let $\omega_1,\omega_2$ be the roots of your quadratic polynomial so that $f(z)= (z-1)(z-w_1)^{-1}(z-w_2)^{-1}$. The radius of convergence will be the least of the distances of $i\sqrt 3$ to the points $w_1,w_2$. Since $z-1$ is a polynomial, it suffices you find the Taylor series of $(z-w_1)^{-1}(z-w_2)^{-1}$. But $$(z-w_1)^{-1}(z-w_2)^{-1}=\frac{1}{w_1-w_2}\left(\frac{1}{z-w_1}-\frac{1}{z-w_2}\right)$$

Hence it suffices we find the Taylor series of $\dfrac 1{w-z}$ around $v$ whenever $v\neq w$. Now $$\frac{1}{w-z}=\frac{1}{(w-v)-(z-v)}=\frac{1}{w-v}\frac{1}{1-\dfrac{z-v}{w-v}}=\frac{1}{w-v}\sum_{\nu \geqslant 0}\left(\frac{z-v}{w-v}\right)^\nu$$

for $|z-v|<|w-v|$.

$\endgroup$
1
  • $\begingroup$ Brilliant. Thank you. We can use the fact that $\frac{1}{w-z}$ is a sum of a geometric series. $\endgroup$ Jan 8, 2015 at 9:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .