Simplifying $\oint_{|z|=1} \frac{-z^4 + 2z^2 - 1}{4i z^3} \, \mathrm dz$

Question

My teacher says that my work below is wrong, and is like claiming that $z = z^4/z^3.$ He insists on a mathematical proof, and I’m unsure how to proceed. I’ve checked many books and references, but none of them indicate that my approach is incorrect.

\begin{align*} \oint_{|z|=1} \left( \frac{1}{2} - \frac{1}{4}(z^2 + z^{-2}) \right) \cdot \frac{\mathrm dz}{iz} &= \oint_{|z|=1} \frac{-z^4 + 2z^2 - 1}{4i z^3} \, \mathrm dz \\ &= 2\pi i \cdot \text{Res}\left( \frac{-z^4 + 2z^2 - 1}{4i z^3}, z=0 \right) \\ &= 2\pi i \cdot \frac{1}{2!} \lim_{z \to 0} \frac{d^2}{dz^2} \left( \frac{-z^4 + 2z^2 - 1}{4i} \right) \\ &= 2\pi i \cdot \frac{1}{2} \cdot \left( \frac{1}{4i} \cdot 4 \right) \\ &= 2\pi i \cdot \frac{1}{2i} = \pi \end{align*}

Answer 1

$$\text{Res}_0 \left(\frac{-z^4+2z^2-1}{4iz^3}\right) = \frac1{4i}\text{Res}_0 \left(-z+2z^{-1}-z^{-3}\right) = \frac{2}{4i} = -\frac12i.$$

You're using the formula for the residue at a third-order pole. Although it is not incorrect, it is a crazy way to do this computation when you can just simplify the algebra, as I did, and read off the $z^{-1}$ coefficient in a second. But I'm on your side (reluctantly); I don't see how your teacher can take off points here, even though your solution is awkward and overly complicated. You even included the "$\lim\limits_{z\to 0}$" and so his pedantic point is out of order. Away from $z=0$, of course $z^4/z^3$ and $z$ are equal.

EDIT: Yes, I didn’t bother to check your original formula. The coefficient $2i$ is in fact wrong. I was concentrating on the method.

Answer 2

...my teacher said my solution is wrong [because] doing this is like claiming $z = z^4/z^3$.

Clearly, over the domain $|z|=1$,

$$\frac{1}{iz} \left( \frac{1}{2} - \frac{1}{4}(z^2 + z^{-2}) \right) = \frac{-z^4 + 2z^2 - 1}{4i z^3}$$

Thus,

$$ \oint_{|z|=1} \frac{1}{iz} \left( \frac{1}{2} - \frac{1}{4}(z^2 + z^{-2}) \right) dz = \oint_{|z|=1} \frac{-z^4 + 2z^2 - 1}{4i z^3} \, dz $$

Whatever happens away from the domain of integration is irrelevant regarding this transformation. There could be a trillion poles at $z=0.42$ for all that it matters.

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The other equalities follow directly from:

2. Cauchy's Residue theorem: $$\oint _{\gamma }f(z)\,dz=2\pi i\sum_k \operatorname {Res} (f, z=c_{k})$$
3. Residue of higher-order poles: $$\operatorname {Res} (f, z=c)={\frac {1}{(p-1)!}}\lim _{z\to c}{\frac {d^{p-1}}{dz^{p-1}}}\left((z-c)^{p}f(z)\right)$$
4. Derivative of polynomial $$\frac{d^2}{dz^2} \sum_k a_k z^k = \sum_k a_k k (k-1) z^{k-2}$$ and limit of continuous function (e.g. polynomial) $$\lim_{z \to c} f(z) = f(c).$$
5. Algebra.

Answer 3

Also, on the unit circle, $z=e^{i\theta}$ and the integral in OP is $$\int_0^{2\pi}\left(\frac12-\frac12\cos2\theta\right)d\theta\\ =\left(\left.\frac12\theta-\frac14\sin2\theta\right)\right\vert_0^{2\pi}=\pi.$$ So no need to complex calculus and teachers.