Separation of variables in ODE with complex conjugate function I am trying to solve the following system:
$$\dot z = \bar z·e^{it}$$
$$z:\mathbb{R}\rightarrow\mathbb{C},\qquad z(t_0) = z_0$$
I cannot figure out how to properly separate the right- and lefthandside. I tried:
$$\int\frac{z}{||z||^2}dz + C= -ie^{it}$$
but wasn't successful.
I will be glad about any tips.
 A: You're not gonna like this, but the answer is fairly convoluted. Divide both sides of the equation by $z$ and rewrite it in the form of $z = e^{\rho + i\phi}$:
$$\frac{\dot{z}}{z} = \frac{\bar{z}}{z}e^{it} \implies \dot{\rho} + i\dot{\phi} = e^{i(t-2\phi)}$$
where $\phi$ and $\rho$ are real functions. Take a look at the imaginary part only to get the ODE
$$\dot{\phi} = -\sin(2\phi - t) \implies \dot{\varphi}+1 = -2\sin\varphi$$
from the substitution $\varphi = 2\phi - t$. Then finagling around with trig identities allows us to integrate the separated variables.
$$-t+c = \int \frac{d\varphi}{1+2\sin\varphi} = \frac{1}{\sqrt{3}}\int \frac{\frac{1}{2}\sec^2 \frac{\varphi}{2}}{\tan \frac{\varphi}{2} + 2 - \sqrt{3}} - \frac{\frac{1}{2}\sec^2 \frac{\varphi}{2}}{\tan \frac{\varphi}{2} + 2 + \sqrt{3}}\:d\varphi$$
$$\implies \frac{\tan \frac{\varphi}{2} + 2 - \sqrt{3}}{\tan \frac{\varphi}{2} + 2 + \sqrt{3}} = Ce^{-\sqrt{3}t} \implies \varphi = 2\tan^{-1}\left(\frac{2\sqrt{3}}{1-Ce^{-\sqrt{3}t}}-(2+\sqrt{3})\right)$$
which means that
$$\phi = \frac{t}{2}+\tan^{-1}\left(\frac{2\sqrt{3}}{1-Ce^{-\sqrt{3}t}}-(2+\sqrt{3})\right)$$
Now we would have to plug this back into the real part of the equation and solve for $\rho$. Trig triangles are your friends. Good luck.

$\textbf{EDIT}$: If the unboundedness of the solution is all that is required, then this solution is very close. For ease of writing, denote
$$\phi = \frac{t}{2} + f(t) \implies \dot{\phi} = \frac{1}{2} + f'(t)$$
Instead of plugging back in to solve for $\rho$, the first equation in this post implies
$$ \dot{\rho}^2 + \dot{\phi}^2 = 1 \implies \dot{\rho^2} = 1 - \left(\frac{1}{2}+f'(t)\right)^2$$
Now take the limit as $t\to\infty$. Notice that $f(t)$ approaches $\tan^{-1}(2\sqrt{3} - (2+\sqrt{3})$ monotonically and smoothly. Thus $f'(t) \to 0$ which means we have that
$$\dot{\rho}^2 \to \frac{3}{4}$$
which means the magnitude $|z|$ will grow without bound.
