I'm searching for a function $\theta(\phi)$ shuch that $\theta(0)=0$, $\theta(\phi_f)=2n\pi$, and
$$ \int_0^{\phi_f} \cos(\theta(\phi)) d \phi=0,\qquad \int_0^{\phi_f} \sin(\theta(\phi)) d \phi = 1 $$ where $\phi_f$ is some angle, and $n$ is some integer value.
They are clearly integral equations, but I find Laplace's method difficult due to the composition. On the other hand I have tried to test different plausible solutions parameterizing the trajectory but condition $\theta(\phi_f)=2n\pi$ is complicated to take into account so I wanted to ask you if you have any idea how to solve this problem?
(Partial answer)
By linearity of the integral, the two integral conditions can be merged into the more simple constraint $$ \int_0^{\phi_f} e^{i\theta(\phi)} \mathrm{d}\phi = i. $$ If we denote by $F(\phi)$ the antiderivative of $f(\phi) = e^{i\theta(\phi)}$, this constraint translates as $F(\varphi_f) - F(0) = i$. Moreover, by definition, we have $F'(\varphi) = e^{i\theta(\phi)}$, hence the additional conditions $F'(\varphi_f) = F'(0) = 1$.
Once you have found a function $F$ sastisfying to these relations, your desired function is given by $\theta_k(\phi) = -i\ln(F'(\phi)) + 2\pi k$, $k \in \mathbb{Z}$. Note that you still need to specify the branch of the logarithm.