1st order separable ODE involving the complex conjugate of the dependent variable Is there a closed form (complex) solution $z(t)$ to the equation
\begin{align}
\frac{dz}{dt}=f(t)\bar{z},
\end{align}
(the bar means complex conjugate) for any given complex valued function $f$ of a real variable $t$? The usual approach to deal with separable equations gives
\begin{align}
\int\frac{1}{\bar{z}}dz=\int{f(t)dt}.
\end{align}
However, the integral on the left is path dependent, so it is apparently not possible to obtain a solution (not even implicit) from this approach.
 A: One cannot give a general answer because a closed form $z(t)$ exists or doesn't exist depending on the kind of function $f(t)$. As shown below, one have to solve a linear system of two équations. This is equivalent to a second order linear ODE. All second order linear ODE cannot be analytically solved and solutions given on closed form. But, in some cases, it is possible depending on the kind of functions involved.
In the particular case of real $f(t)$ , a closed form is obtained insofar an antiderivative of $f(t)$ is known on a closed form.
 
A: Here is an alternative approach.
When you are given the equation
$$z' = f(t) \overline{z},$$
 you are also implicitly given
$$\overline{z}' = \overline{f}(t) z$$
by conjugation.
So let $Z =(z,\overline{z})$, then we have
$$Z' = \left(\begin{matrix}0&f(t)\\ \overline{f}(t)&0
\end{matrix}\right)Z
$$
This is $2$d linear differential equation with time dependant coefficients and it is well established that no general closed form solution exists.
A series solution does exist however, in the form of a Magnus Expansion; ie
$$Z(t) = \exp(\Omega(t))Z(0) $$
where 
$$\Omega(t) = \sum_{k=1} \Omega_k(t)$$
Each $\Omega_k$ is expressed in term of integrals over $k-1$ nested commutators. You can read about this yourself, but i would just like to consider the case where $f(t)$ is real.
The first term is given by
$$\Omega_1(t) = \int_0^t\left(\begin{matrix}0&f(\tau)\\ \overline{f}(\tau)&0
\end{matrix}\right)d\tau$$
and the second by
$$
\Omega_2(t) =\int_0^t\int_0^\tau
\left(\begin{matrix}f(s)\overline{f}(\tau)-f(\tau)\overline{f}(s)&0\\0& f(\tau)\overline{f}(s)-f(s)\overline{f}(\tau)
\end{matrix}\right) ds d\tau.
$$
From this we see that if $f$ is real, $\Im[f(\tau)\overline{f}(s)]=0$ for all $\tau, s$ and hence $\Omega_2(t) =0$ (as does $\Omega_k=0$ for $k>2$). Once you put this all together you get the same thing as what JJacquelin posted.
