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I want to find the the curve integral

$$\int_γ \frac{1}{1 - z + z^2 - z^3 } dz$$

with $γ$ passing through the following sets counter-clockwise once.

a) $\{z \in \mathbb{C}, |z - i| = 1\}$

b) $\{z \in \mathbb{C}, |z + i| = 1\}$

c) $\{ z \in \mathbb{C}, |z| = 2\}$

(I want to evaluate the integral for each set seperately, so we have three different $\gamma$'s, not one curve going through all of them at once.)

Now I'm having trouble on how to determine the integral. I first tried figuring out the three sets. I think the set in a) is the unit circle, but centered around $i$, and the set in b) is the unit circle, centered around $-i$. And c) seems to simply be a circle around $0$ with a radius of $2$.

So I think I could solve this by parameterizing $\gamma$. We have that $e^{it}$ would go around the unit circle counter-clockwise, so I think that $\gamma_{a)} = e^{it} + i$ would for example parameterize the set in a). But how do I continue from there? Do I need to integrate $\frac{1}{1-z+z^2-z^3}$ entirely (because it apparently isn't very nice to integrate), or is there a shortcut or trick?

For example, I thought about using the Cauchy-integral formula (which would, if applicable, give us that the integral is $0$), but I don't think it's applicable here, because for e.g. $z = i$, we have that $1 - z + z^2 - z^3 = 1 - i + (-1) - (-i) = 0$, so the function isn't defined there and we have a hole in the domain the function is defined on. (And it frankly seems to be too easy if one could just say that the integral is $0$, so I don't think it's the case.)

Still, I'm trying to figure out if there are other theorems one could apply, that help us simplifying the expression. The way this task was given to me made it seem like there's a much easier way; one that doesn't involve calculating all three integrals "on foot"; but I can't think of any right now.

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  • $\begingroup$ Where are the poles (zeroes of $1-z+z^2-z^3$) located ? Draw these point on a plane together with the circles. Hint : the circle do not pass through any of the poles. Cauchy's residue theorem applies easily here. $\endgroup$
    – Tom-Tom
    Nov 17, 2015 at 14:10
  • $\begingroup$ It may be helpful to note that $$ 1 - z + z^2 - z^3 = \frac{1 - z^4}{1 + z} $$ $\endgroup$
    – Ben Grossmann
    Nov 17, 2015 at 14:12
  • $\begingroup$ Ok, so we have that $1 - z + z^2 - z^3 = 0$ for $z = 1, -1, i, -i$, which must be the only roots. They indeed don't interfere with the other circles. I must admit though that I didn't have Cauchy's residue theorem yet in my class, so I'm probably not supposed to use it (and I don't really know what it states either). I will look it up nevertheless and try to understand it, but would there be another trick or theorem that could help? $\endgroup$
    – moran
    Nov 17, 2015 at 14:30
  • $\begingroup$ 3 roots, is not -1 $\endgroup$
    – Nosrati
    Nov 17, 2015 at 14:38
  • $\begingroup$ Oh, I see. Thanks for pointing that out; can't edit my comment anymore. The degree of the polynomial should have alerted me. $\endgroup$
    – moran
    Nov 17, 2015 at 14:40

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