$\oint_{\gamma}(2z-3\bar z+1)\,dz$ where $\gamma$ is the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ $\oint_{\gamma}(2z-3\bar z+1)\,dz$ 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.
 A: 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.
A: $$ z(t) = 2 \cos t + i3 \sin t$$
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 $t$
