Note this question is a part of a bigger question, I have done a) and b).
Problem 2
Let $u(x,y) = x e^x \cos y - y e^x \sin y $
a) Show that u is harmonic in the entire plane
b) Find a function $v$ such that $f = u + i v$ is analytic
c) Show that $$\int_0^{2\pi} \cos \theta e^{\cos\theta} \cos(\sin \theta) - \sin \theta e^{\cos\theta} \sin(\sin \theta)\,\mathrm{d}\theta=0$$
For a) one can note that $\text{Re}(ze^z) = u(x,y)$, hence $u$ is harmonic.
Or show that $\nabla^2 u(x,y)=\partial u = \partial^2u/\partial x^2 + \partial^2 u/\partial x^2 =0$.
Part b) can be found by either taking $v=\text{Im}(z e^z) =e^x(x\sin y+y\cos y)$, or integrate the Cauchy-Riemann equations.
I have two "proofs" for c) and I am quite unsure about the finer works in both of them. Note that if $h(\theta)$ denotes the integrand then $u(\cos \theta,\sin \theta)=h(\theta)$.
We know that there exists a function $f$ such that $f = u + iv$, and $f$ is analytic.
The path integral over some closed curve is zero, over an analytic function. So the following should hold
$$
\oint_C f\,\mathrm{d}s = \oint_C u\,\mathrm{d}s + i \oint_C v\,\mathrm{d}s = 0\,,
$$
where $C$ is some closed curve. This means that
$$
\oint_C u\,\mathrm{d}s = \frac{1}{i} \oint_C v
$$
What I want to say is that both line integrals are reall, so the left handside
is imaginary and the right handside is real. My argument for this is that both $u$ and $v$ lives in $C^\infty$ and hence has no singularities in $\mathbb{C}$.
Is this correct? Is there any better argument that both integrals are real?
Assume that the integrals are real, then the only possibility for the equation to hold is if both integrals are zero. if we take $C$ to be the unit circle, then we can parametrize the integral over u
$$
\oint_{|z|=1} u\,\mathrm{d}s = \int_0^{2\pi} u(e^{i \theta})|i e^{i \theta}| \mathrm{d}\theta = \int_0^{2\pi} h(\theta)\,\mathrm{d}\theta = 0
$$
Not quite sure how I should parametrize the integral such that it becomes real. I do not quite see why $u(e^{i\theta})$ equals $h(\theta)$, the first integral lives in $\mathbb{C}$ and the second in $\mathbb{R}$.
Could I instead have integrated $z = u + i v$ in $\mathbb{R}^2$ instead of $\mathbb{C}$
since they are isomorphic? It seems somewhat strange, but atleast then the parametrization of $\oint u$ will be real.
My proffesor said I could use the maximul value principle to evalute this integral.
But again this integral lives in $\mathbb{C}$ and $h(\theta)$ lives in \mathbb{R}^2.
Otheriwise I see why that principle works, since $u(0,0)=0$.
Any inputs or suggestions for making my argument work, or a similar argument would be much appreaciated. I have spent a few days pondering over this problem.
Answer
Variant 1: Harmonic functions have the mean value property,
$$f(z) = \frac{1}{2\pi} \int_0^{2\pi} f(z + re^{i\varphi})\,d\varphi$$
if $f$ is harmonic in $\Omega$ and $\overline{D_r(z)} \subset \Omega$.
$u$ is an entire harmonic function, hence
$$u(0) = \frac{1}{2\pi}\int_0^{2\pi} u(e^{i\varphi})\,d\varphi = \frac{1}{2\pi}\int_0^{2\pi} \cos(\varphi)e^{\cos\varphi}\cos (\sin\varphi) - \sin(\varphi)e^{\cos\varphi}\sin(\sin\varphi)\,d\varphi.$$
Whether we write $z$ and $re^{i\varphi}$ or $(x,y)$ and $(r\cos \varphi, r\sin\varphi)$ is completely immaterial. The complex notation is just more convenient sometimes.
Variant 2: Consider the analytic function $f = u+iv$.
Since $u$ and $v$ are both real, and $d\varphi$ is also real, we have
$$\begin{align}
\int_0^{2\pi} u(\cos\varphi,\sin\varphi)\,d\varphi &= \operatorname{Re}\left(\int_0^{2\pi} u(\cos\varphi,\sin\varphi)\,d\varphi + i\int_0^{2\pi} v(\cos\varphi,\sin\varphi)\,d\varphi\right)\\
&= \operatorname{Re} \int_0^{2\pi} f(e^{i\varphi})\,d\varphi.
\end{align}$$
Now,
The path integral over some closed curve is zero, over an analytic function.
is not correct as stated. On the one hand, the closed curve must not wind around any point in the complement of the function's domain - but since we have an entire function, that is vacuously satisfied here. More pertinent in the case at hand is that the integral theorem concerns only integrals with respect to $dz$ (it's a theorem about holomorphic differential forms), but here the integrand is $f(z)\,d\varphi$, not $f(z)\,dz$. Thus Cauchy's integral theorem does not apply.
However, for integrals over a circle, we have a simple correspondence between $dz$ and $d\varphi$. If we parametrise the circle as $\gamma(\varphi) = z_0 + r e^{i\varphi}$, then we have
$$dz = \gamma'(\varphi)\,d\varphi = ire^{i\varphi}\,d\varphi = i(z-z_0)\,d\varphi,$$
so we get
$$\int_0^{2\pi} f(e^{i\varphi})\,d\varphi = \int_{\lvert z\rvert = 1} f(z)\frac{dz}{iz},$$
and we see that that leads to Cauchy's integral formula,
$$\frac{1}{i} \int_{\lvert z\rvert = 1} \frac{f(z)}{z}\,dz = 2\pi\: f(0).$$
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