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$\oint_{\gamma}(2z-3\bar z+1)\,dz$ where the contour $\gamma$ is the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ traversed clockwise.

My difficulty is how to translate the ellipse formula to analytical path . for circle it's easy ($e^{zi}$), but I don't know how to do so for ellipse.

Mark Viola
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  • it is difficult to construct explicitly an analytic mapping from the circle to the ellipse (elliptical integrals) so i suggest doing it by a smooth parameterization and split into real and imaginary parts $x=2\cos t, y=3 \sin t, dz=dx+idy=(x'+iy')dt$ etc; note that only the $\bar z dz$ integral requires computation as the rest are zero by Cauchy – Conrad Dec 03 '20 at 18:01
  • What's the meaning of $z^-$? – José Carlos Santos Dec 03 '20 at 18:04
  • @JoséCarlosSantos The OP wrote "the $-3z$ should be $z$ bar. – Mark Viola Dec 03 '20 at 18:07
  • @MarkViola Thank you. I missed that. – José Carlos Santos Dec 03 '20 at 18:08
  • $\int (2z + 1) , dz = 0$ due to the Cauchy integral theorem. And $\int \bar z , dz$ is related to the enclosed area: https://math.stackexchange.com/q/445781/42969. – Martin R Dec 03 '20 at 18:10
  • Is it cheating if we claim the integral of $\overline z dz$ around a ccw contour is $2i$ times the enclosed area? – Oscar Lanzi Dec 03 '20 at 19:19

2 Answers2

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Forget explicit parameterization of $\gamma$, just use Stoke's theorem. In particular, use the version stated in complex coordinates.

Let $E$ be the ellipse bounded by $\gamma$. Since $\gamma$ walks around $E$ in clockwise direction, it is "negative" to the orientation of $\partial E$, the boundary of ellipse. Apply Stoke's theorem in complex coordinates, we have

$$\int_\gamma (2z - 3\bar{z} +1 ) dz = \int_{-\partial E}(2z - 3\bar{z} + 1) dz = -\int_E d(2z - 3\bar{z} + 1) \wedge dz\\ = 3\int_E d\bar{z} \wedge dz = 6i \int_E \frac{d\bar{z}\wedge dz}{2i}$$ In terms of Cartesian coordinates,

$$\frac{d\bar{z}\wedge dz}{2i} = \frac{d(x-iy) \wedge d(x+iy)}{2i} = dx \wedge dy$$ is simply the area element. Since ellipse $E$ has semi-major/minor axes $3$ and $2$, we have:

$$\int_\gamma (2z - 3\bar{z} +1 ) dz = 6i\verb/Area/(E) = 6i(6\pi) = 36\pi i$$

For comparison, let us redo the computation in Cartesian coordiantes.

We can parametrize $E$ as

$$[0,2\pi] \ni \theta\quad\mapsto\quad (x,y) = (2\cos\theta,\color{red}{-}3\sin\theta) \in \mathbb{R}^2 \sim \mathbb{C}$$

Since $\gamma$ walks around $E$ in clockwise direction, the sign in front of $\sin\theta$ is negative instead of positive. Plug these into original integral, it becomes

$$\begin{align} &\int_0^{2\pi} (2(2\cos\theta - 3\sin\theta i) - 3(2\cos\theta + 3\sin\theta i) + 1)(-2\sin\theta - 3\cos\theta i) d\theta\\ = &\int_0^{2\pi} -(2 + 41\cos\theta)\sin\theta + (30\sin^2\theta + 6\cos^2\theta - 3\cos\theta)i d\theta\end{align}$$ Throwing away terms which clearly don't contribute, we get

$$\begin{align}\int_\gamma(2z - 3\bar{z} +1 )dz &= i\int_0^{2\pi}(30\sin^2\theta + 6\cos^2\theta)d\theta\\ &= i(30\pi + 6\pi) = 36\pi i\end{align} $$ Same number $36\pi i$ we obtained before.

achille hui
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0

$$ z(t) = 2 \cos t + i3 \sin t$$

Explanation:

$$ z(t) = x(t) + i y(t)$$

Now, for the ellipse for a trignomeetric parameterization: $x=2 \cos t$ $y=3 \sin t$, plug that same thing into the complex function in $t$

tryst with freedom
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