The number of complex numbers z satisfying z^6=bar z and Re(z)>0 is

Taking moduli, $|z|^6=|z|$, so $|z|=0$ or $|z|=1$. The value $z=0$ is excluded by $\operatorname{Re}(z)>0$. For $|z|=1$ we have $\bar z=1/z$, hence $z^{7}=1$ and $z=e^{2\pi i k/7}$ for $k=0,1,\dots,6$. The real part is $\cos(2\pi k/7)$. It is positive when $2\pi k/7$ lies within $(-\pi/2,\pi/2)$ modulo $2\pi$, i.e. $k/7<1/4$ or $k/7>3/4$, which gives $k=0,1,6$ only (for $k=2$, $\cos(4\pi/7)<0$; $k=5$ is symmetric to $k=2$). So exactly three roots qualify. A direct numerical check of all seven roots confirms that only $k=0,1,6$ have positive real part.