I have a problem with a few parts (possibly related to each other) that I'm not sure about:

Show that $f(z) = z/(z^2 - 4z +1)^2$ is holomorphic except at two points, $\alpha$ and $\beta$ with $|\alpha| < 1 < |\beta|$

This part I'm OK with, since you just find the two roots of the polynomial on the bottom, no problem.

Show that the coefficients $c_n$ of the Taylor expansion of $g(z) = z/(z-\beta)^2$ about $\alpha$ are: $$c_n = \frac{\alpha + n\beta}{(\beta-\alpha)^{n+2}}$$

This part is not too difficult either, I think, since I can just use partial fractions and observe that all of the terms are slightly modified geometric series, the answer falls out after some work.

The next part is giving me trouble:

With reference only to Taylor's theorem, evaluate $$ \int_\gamma \frac{z\;dz}{(z^2 - 4z +1)^2}$$ (where $\gamma$ denotes the unit circle centre at 0) and hence show that $$\int^{2\pi}_0 \frac{d\theta}{(2-\cos \theta)^2} = \frac{4\pi}{3\sqrt3}$$

The emphasis is mine. I guess it means it wants us to use the series we just worked out. I can't see how to use the result to calculate the integral. I can sort of see one way to do it using integration by parts and Cauchy's formula:

\begin{align} \int_\gamma \frac{z\;dz}{(z^2 - 4z +1)^2} &= \int_\gamma g(z) \frac{1}{(\alpha-z)^2} dz \\ &= \int_\gamma g(z) \left(\frac{d}{dz}\frac{1}{\alpha-z}\right) dz \\ &= -\int_\gamma g'(z) \frac{1}{\alpha-z} dz \;\;\mbox{(this is by parts)} \\ &= \int_\gamma \frac{g'(z)}{z-\alpha} dz \\ &= 2\pi i g'(\alpha) \;\;\mbox{(this is by Cauchy's formula)} \\ &= \frac{4\pi i}{\sqrt3} \end{align}

Are all of these steps valid? I think so, since $g(z)$ is holomorphic inside $\gamma$, so I can't see any failed condition. But my problem here is that I haven't used the earlier part at all.

I realise for the second part, I'm probably supposed to use the substitution $z = e^{i\theta}$ and put it through the first integral, but I'm having trouble putting it through since I can't see anywhere I'll be able to pull out $1/3$ to relate it to my previous answer.

I would appreciate if someone pointed out my mistakes.

  • $\begingroup$ the taylor expansion you have calculated, around which point is it? $\endgroup$ – M. Van Nov 12 '16 at 21:42
  • $\begingroup$ Sorry, it's about $\alpha$. I just edited it in. $\endgroup$ – pizzaroll Nov 12 '16 at 21:44

Write $$\frac{z}{(z^2-4z+1)^2}=\frac{z}{(z-\alpha)^2(z-\beta)^2}.$$

The radius of convergence of the taylor series of $\frac{z}{(z-\beta)^2}$ about $\alpha$ is the distance between $\alpha$ and $\beta$. If the ball $B(\alpha, |\alpha-\beta|)$ contains $\gamma$ entirely(this can be checked by explicitly calculating $\alpha$ and $\beta$), then one can write, with help of the taylor series, and by the fact that the sequence of functions $(\sum_{k=0}^n c_k(z-\alpha)^{k-2})$ converges uniformly on $B(\alpha, |\alpha-\beta|)-\{\alpha\}$, we have

$$\int_\gamma \frac{z}{(z-\alpha)^2(z-\beta)^2} dz=\int_{\gamma}\frac{1}{(z-\alpha)^2} \sum_{k=0}^{\infty} c_k (z-\alpha)^k dz=\int_{\gamma} \sum_{k=0}^{\infty}c_k(z-\alpha)^{k-2}dz=\sum_{k=0}^{\infty} \int_{\gamma}c_k(z-\alpha)^{k-2}dz=c_12\pi i,$$ since all other terms have a primitive function on $\mathbb{C}-\{\alpha\}$. I hope this helps.


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