# Evaluate $I=\int_{\gamma (0,1)}\frac{zdz}{(z^2-4z+1)^2}$ using Taylor's Theorem

I've shown that $f(z)=\frac{z}{(z^2-4z+1)^2}$ is holomorphic apart from at points $\alpha=2-\sqrt3$ and $\beta=2+\sqrt3$ and that the talyor coefficient of $g(z)=\frac{z}{(z-\beta)^2}$ centred at $\alpha$ is $$c_n=\frac{\alpha+n\beta}{(\beta-\alpha)^{n+2}}$$

Now I'm not sure how to use Taylor's theorem to evaluate the integral I, where $\gamma (0,1)$ is the unit circle centred at the origin Thanks!

I can't seem to comment, but no, I don't think we've learnt residues. I have Cauchy's integral formula if that's similar?
Thank you so much for taking the time to answer my question. It's much much clearer now. I can't seem to upvote or select an answer though, any idea why this is?

• Do you know about residues? Nov 18, 2013 at 22:37
• I read your response in the middle of typing my answer, so I decided to finish it anyway even though it uses residues Nov 18, 2013 at 22:55
• Updated with answer using only Taylor's Theorem. Nov 18, 2013 at 23:53
• I have no idea why you can't upvote or accept my answer, I will flag the question for moderator attention so they can look into it. Nov 19, 2013 at 22:47
• You can't upvote answers because your account is not registered (though you should still be able to accept answers -- if you can't do this, something is surely wrong). To register your account I believe you go to your user profile, click the "my logins" tab near the top, and then add a login. Nov 20, 2013 at 5:40

This answer assumes $\gamma(0,1)$ is $\gamma\colon [-\pi, \pi]\to \Bbb C, \theta \mapsto e^{i\theta}$.

The Residue theorem yields $$\displaystyle\int_\gamma f=2\pi i\left(\operatorname{Res}(f,\alpha)+\operatorname{Res}(f,\beta)\cdot 0\right)=2\pi i\operatorname{Res}(f,\alpha).$$

All that is left to find is $\operatorname{Res}(f,\alpha)$. You should know that $\operatorname{Res}(f,\alpha)$ is the coefficient of $(z-\alpha)^{-1}$ in the Laurent series of $f$ in a set like $\{z\in \Bbb C\colon 0<|z-\alpha|<r\}$, for some $r>0$.

You've already found, for all $z$ in the set above, $\displaystyle \dfrac z{(z-\beta)^2}=\sum \limits_{n=0}^{+\infty}\left(c_n(z-\alpha)^n\right)$, thus

$$\dfrac{z}{(z^2-4z+1)^2}=\dfrac z{(z-\alpha)^2(z-\beta)^2}=\sum \limits_{n=0}^{+\infty}\left(c_n(z-\alpha)^{n-2}\right).$$

Therefore the coefficient of $(z-\alpha)^{-1}$of $f$ in its Laurent series is $c_1\color{grey}{=\dfrac{4}{(2\sqrt 3)^3}=\dfrac{1}{6\sqrt 3}}.$ Which agrees with Wolfram Alpha.

Finally $\displaystyle\int_\gamma f=2\pi i\cdot \dfrac{1}{6\sqrt 3}=\dfrac{\pi i}{3\sqrt 3}$.

In this part of the answer I'm mainly following theorem 5.2.1 in this notes.

Taylor's theorem (or perhaps just its proof, depending on the exact statement), says that in the series of $\varphi$ around $z_0$, $$\color{grey}{\varphi (z)=}\sum \limits_{n=0}^{+\infty}\left(a_n(z-z_0)^n\right),$$ for each $n\in \Bbb N_0$, $a_n$ is given by $$\displaystyle \color{grey}{a_n=}\dfrac 1{2\pi i}\int_{C_r}\dfrac{\varphi (z)}{(z-z_0)^{n+1}}\mathrm dz.$$

Setting $n=1,\varphi =g$ and $z_0=\alpha$ gives us $\displaystyle \int _{C_1}f=2\pi i\cdot c_1=\dfrac{\pi i}{3\sqrt 3}$.