= Solution
The <conformal map> $\zeta=z^3$ sends the wedge $0<\arg z<\pi/3$ to the upper half-plane. The physical <vortex> maps to $\zeta=c^3$; an opposite image <vortex> at $\overline c^{,3}$ makes the real axis a streamline. Pulling this potential back gives
$$
w(z)=-\frac{i\Gamma}{2\pi}\log(z^3-c^3)
+\frac{i\Gamma}{2\pi}\log(z^3-\overline c^{,3}).
$$
Since $z^3-c^3=(z-c)(z^2+cz+c^2)$, splitting the first logarithm, up to an additive constant, gives exactly the required form. The image construction enforces zero normal velocity on both rays.
To find the <vortex> motion, hold $c$ fixed while differentiating with respect to $z$, remove the self-induced singular term $-i\Gamma/(2\pi(z-c))$, and then evaluate at $z=c$. The remaining <complex velocity> is
$$
\frac{i\Gamma}{2\pi}\left[-\frac{2c+c}{c^2+c^2+c^2}
+\frac{3c^2}{c^3-\overline c^{,3}}\right].
$$
A <complex velocity> $u-iv$ equals $d\overline c/dt$, rather than $dc/dt$. Thus
$$
\boxed{\frac{d\overline c}{dt}=\frac{i\Gamma}{2\pi}
\left(-\frac1c+\frac{3c^2}{c^3-\overline c^{,3}}\right).}
$$
The overbar on the source derivative is essential to the trajectory calculation.
Write $c=re^{i\theta}$. Since $c^3-\overline c^{,3}=2ir^3\sin(3\theta)$, multiply the equation by $e^{i\theta}$ to obtain
$$
\dot r-ir\dot\theta
=\frac{3\Gamma}{4\pi r}\cot(3\theta)+\frac{i\Gamma}{4\pi r}.
$$
Taking real and imaginary parts gives
$$
\dot r=\frac{3\Gamma}{4\pi r}\cot(3\theta),\qquad
\dot\theta=-\frac{\Gamma}{4\pi r^2}.
$$
Therefore
$$
\frac d{dt}\big(r\sin(3\theta)\big)
=\dot r\sin(3\theta)+3r\cos(3\theta)\dot\theta=0,
$$
proving the trajectory invariant
$$
\boxed{r\sin(3\theta)=\text{constant}.}
$$
The calculation also covers $\Gamma=0$, when the <vortex> position is stationary.
This is the sixty-degree case of a <point vortex in a wedge>.
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