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I'm trying to solve the following problem:

For what simple piecewise smooth loops $\gamma$ does the following equation hold: $$\int\limits_\gamma \frac{1}{z^2 + z + 1} \, \mathrm{d}z= 0$$

I'm allowed to appeal to Cauchy's integral theorem, and I have determined that the roots of the denominator are $\frac{-1}{2} \pm \sqrt{3}i$. But I am not exactly sure how to proceed, or even what the ideal solution would be. For example, I know that the integral is $0$ over any loop that is homotopic to a constant loop (at a point in the domain of the function), but I'm not sure of what else I should say.

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  • $\begingroup$ I think you just need to find a closed simple path that doesn't include those roots, no? $\endgroup$ Commented Mar 21, 2016 at 0:21
  • $\begingroup$ @LoveTooNap29 Yes, I guess I just thought that that would be too obvious... :P $\endgroup$
    – User
    Commented Mar 21, 2016 at 0:28

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$z^2+z+1 = \left( z^2+z+\frac 1 4 \right) + \frac 3 4 = \left( z + \frac 1 2 \right)^2 + \frac 3 4 = 0$ if and only if $z= -\frac 1 2 \pm i\sqrt{\frac 3 4} = - \frac 1 2 \pm i\frac{\sqrt 3} 2$, so you're missing a $2$.

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

Fortunately you said the path is "simple" so that precludes winding $10$ times around one of these points and $7$ times around the other, etc. But you didn't say "closed", i.e. returning to its starting point.

There is a simple pole at each of those two points. If a curve winds once around one of those points and not around the other, then the integral will not be $0$. If it winds around neither, the integral will be $0$.

But what if it winds around both, once each? Then the integral will be the sum of the two numbers you get by winding around each once. Will that be $0$, or not?

First suppose $\gamma$ winds once around $- \frac 1 2 + i \frac{\sqrt 3} 2$ and not around $- \frac 1 2 - i \frac{\sqrt 3} 2$. Let $w= z - \left( - \frac 1 2 + i \frac{\sqrt 3} 2 \right)$ so that $dw=dz$. Then we have $$ \int_\gamma \frac 1 { \left( z+ \frac 1 2 + i \frac{\sqrt 3} 2 \right) \left( z+ \frac 1 2 - i \frac{\sqrt 3} 2 \right) } \, dz = \int_{\gamma - \left( - \frac 1 2 + i \frac{\sqrt 3} 2 \right)} \frac{dw}{(w + i\sqrt 3)w} $$ and the curve $\gamma - \left( - \frac 1 2 + i \frac{\sqrt 3} 2 \right)$ winds once around $0$ and not around $-i\sqrt 3.$ The value of $w + i\sqrt 3$ at $w=0$ is $i\sqrt 3$, so the integral is $$ i\sqrt 3 \int_{\gamma - \left( - \frac 1 2 + i \frac{\sqrt 3} 2 \right)} \frac {dw} w. $$ Evaluate that integral and do the same with a curve winding only around the other point and see if they add up to $0$ or not.

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  • $\begingroup$ Oops, sorry, I have edited the question above to say 'loops'. So yes, they are meant to be closed paths. I suppose that will change your answer? $\endgroup$
    – User
    Commented Mar 21, 2016 at 1:09
  • $\begingroup$ @User : No, my answer only considered closed curves. $\qquad$ $\endgroup$ Commented Mar 21, 2016 at 2:04
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The roots of $z^{3}-1$ are $e^{k(2\pi i/3)}$ for $k=0,1,2$ and $$ z^3-1 = (z-1)(z^2+z+1). $$ So $(z-e^{2\pi i/3})(z-e^{-2\pi i/3})=z^2+z+1$, and $$ \frac{1}{(z-e^{2\pi i/3})(z-e^{-2\pi i/3})}=\frac{A}{z-e^{2\pi i/3}}+\frac{B}{z-e^{-2\pi i/3}}, $$ where $A$, $B$ are easily seen to satisfy $A=-B$. Assuming $\gamma$ does not pass through the points $e^{\pm 2\pi i/3}$, you get zero iff the curve $\gamma$ has a winding number around $e^{-2\pi i/3}$ that is the same as the winding number around $e^{2\pi i/3}$.

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