Let $z=\rho e^{i\theta}$ be a complex number. What is the argument of $1-\bar{z}$ in terms of $\theta$? Is it $\pi-\theta$ because the argument of $-\bar{z}$ is $\pi-\theta$?
Follow-up: Is there a way to simplify the argument: $\arctan\left(\frac{\rho\sin\theta}{1-\rho\cos\theta}\right)$?
Since $\tan\left(-\frac{\theta}{2}\right)=\frac{\sin{(-\theta)}}{1+\cos{(-\theta)}}=\frac{-\sin\theta}{1+\cos\theta}=\frac{-\sin\theta}{1-(-\cos\theta)}$
if $\rho=-1$, you would have $\frac{\rho\sin{\theta}}{1-\rho\cos\theta}$ which is the form you are looking for. I don't know if there is any other way to reduce for other values of $\rho$, but in general, i believe it is as far as you can go.