Solution of a differential equation stays on the unit circle once it visits the unit circle Consider the equation (for $\lambda \in C([0,\infty))$) given by;
$\partial_t g_t(z) = g_t(z) \frac{e^{i\lambda (t)} + g_t(z)}{e^{i \lambda (t)} - g_t(z)} $; $g_0(z) =z$.
Here $t \in [0,\infty)$ and $z \in \mathbb{C}$. 
I want to prove that when $z$ is on the unit circle, $g_t(z)$ is also on the unit circle. 
So far, since $g_0(z) = z$ I have that $g_t(z)$ starts on the unit circle when $z$ is in the unit circle. I also see that $\partial_t \log|g_t(z)| = Re \Big( \frac{e^{i\lambda (t)} + g_t(z)}{e^{i \lambda (t)} - g_t(z)} \Big)$
 A: Since you already know that $\partial_t\log|g_t(z)| = Re\left(\frac{e^{i\lambda (t)} + g_t(z)}{e^{i \lambda (t)} - g_t(z)} \right)$, all you need to do is to show that this real part is zero if you start on the unit circle. Divide and multiply by the complex conjugate of the denominator and you get:
$$\frac{e^{i\lambda (t)} + g_t(z)}{e^{i \lambda (t)} - g_t(z)} \times\frac{e^{-i \lambda (t)} - g_t^*(z)}{e^{-i \lambda (t)} - g_t^*(z)}
=\frac{1-|g|^2+2iIm\left(g e^{-i\lambda}\right)}{1+|g|^2-2Re\left(ge^{-i\lambda}\right)}
$$
If you start on the unit circle, i.e. $|z|=1$, you get $|g_0(z)|=1$. Pluging $|g|=1$ in the above equation, you get 
$$
\partial_t\log|g_t(z)| =Re\left( \frac{iIm\left(g e^{-i\lambda}\right)}{1-Re\left(ge^{-i\lambda}\right)}\right)=0
$$
which keeps you on the unit circle.
Edit: Let me clarify the strategy: to show that points on the unit circle stay on the unit circle, I have shown that $|g|=1$ satisfies the differential equation $$
\frac{\partial_t |g|}{|g|} = \frac{1-|g|^2}{1+|g|^2-2Re\left(ge^{-i\lambda}\right)}
$$
