2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I think you just need to find a closed simple path that doesn't include those roots, no? $\endgroup$ – Nap D. Lover Mar 21 '16 at 0:21
  • $\begingroup$ @LoveTooNap29 Yes, I guess I just thought that that would be too obvious... :P $\endgroup$ – User Mar 21 '16 at 0:28
2
$\begingroup$

$z^2+z+1 = \left( z^2+z+\frac 1 4 \right) + \frac 3 4 = \left( z + \frac 1 2 \right)^2 + \frac 3 4 = 0$ 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.

$\endgroup$
  • $\begingroup$ 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 Mar 21 '16 at 1:09
  • $\begingroup$ @User : No, my answer only considered closed curves. $\qquad$ $\endgroup$ – Michael Hardy Mar 21 '16 at 2:04
0
$\begingroup$

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 $e^{2\pi i/3}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.