$3$ mice in the corners of an equilateral triangle crawling at each other ODE problem At each corner of an equilateral triangle, a mouse is positioned. At time $t = 0$ the mice begin crawling towards each other. Mouse $1$ always crawls directly towards mouse $2$, mouse $2$ towards mouse $3$ and mouse $3$ towards mouse $1$. The speed of each mouse is equal to the vector of its position to the position of the mouse they are crawling towards.

What does this last sentence mean?

Determine the trajectories of the mouse by creating a system of differential equations and solving it. What happes for $t \to +\infty$?
Note: It is advantageous to describe the position of the $k$-th mouse by a complex number $z_k$, $k = 1,2,3$. A skilful choice of the origin of the coordinate system is also advantageous
My try:
$z_1(t_0)=i,\ z_2(t_0)=\frac{\sqrt 3}{2}-\frac{1}{2}i,\ z_3(t_0)=\frac{-\sqrt 3}{2}-\frac{1}{2}i$
$z_2(t)=z_1(t)e^{\frac{4\pi}{3}i}=z_1(t)(-\frac{1}{2}-\frac{-\sqrt 3}{2}i)$ (this is a rotation by $240°$)
I'm not sure sure if I can set the ODE like this:
$z_1'(t)=z_2(t)-z_1(t)=z_1(t)(-\frac{3}{2}-\frac{-\sqrt 3}{2}i)$
Therefore $z_1(t)=e^{(-\frac{3}{2}-\frac{3}{2}i)t}c$ where $c=z_1(0)=i$.
And
$\lim_{t\to\infty}z_1(t)=\lim_{t\to \infty}e^{-\frac{3}{2}t}(...)=0$

Let $\omega:=e^{2\pi i/3}$. The three mice always form an equilateral triangle. Therefore we may write
$$z_k(t)=\omega^k \>z(t),\qquad z(0)=1\ .$$
According to the last sentence we have
$$\dot z_k(t)=z_{k+1}(t)-z_k(t)=(\omega-1)z_k(t)\ ,$$
so that we obtain the IVP
$$\dot z(t)=(\omega-1)\>z(t),\qquad z(0)=1$$
with solution
$$z(t)=e^{(\omega-1)t}=e^{-3t/2}\>e^{i\sqrt{3}\>t/2}\ .$$
The limit when $t\to\infty$ is then obvious.
