= Solution
The <conformal map> $\zeta=z^3$ sends the wedge to the upper half-plane. A <fluid vortex> at $c^3$ there has an opposite image at $\bar c^3$, so the pulled-back potential is
$$
w(z)=\frac{i\Gamma}{2\pi}\bigl(-\log(z^3-c^3)+\log(z^3-\bar c^3)\bigr).
$$
The two boundary rays map to the real axis, where the imaginary part is constant, giving the required impermeability. Factoring $z^3-c^3=(z-c)(z^2+cz+c^2)$ gives the printed form up to branch constants.
The <fluid vortex> moves with the regular part of the <complex velocity>, excluding its own singular term. Since $w'=u-iv$, its motion is the conjugated <derivative>. Differentiating the remaining terms and setting $z=c$ gives
$$
\boxed{\frac{d\bar c}{dt}=\frac{i\Gamma}{2\pi}\left(-\frac1c+\frac{3c^2}{c^3-\bar c^3}\right).}
$$
The overbar on the time <derivative> is essential.
Write $c=re^{i\theta}$. Using $c^3-\bar c^3=2ir^3\sin3\theta$ and multiplying the velocity equation by $e^{i\theta}$ gives
$$
\dot r-ir\dot\theta=\frac{\Gamma}{4\pi r}(3\cot3\theta+i).
$$
Thus $\dot r=3\Gamma\cot3\theta/(4\pi r)$ and $\dot\theta=-\Gamma/(4\pi r^2)$. Direct differentiation now yields
$$
\frac d{dt}(r\sin3\theta)=\dot r\sin3\theta+3r\dot\theta\cos3\theta=0.
$$
Hence $\boxed{r\sin(3\theta)=\text{constant}}$ along the trajectory while it stays inside the wedge.
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