Complexity! (3)

It is given that $z = |z|e^{i\theta} = e^{i\theta}$ $\implies \bar z = e^{-i\theta}$ . Then we have:

$\begin{aligned} \frac {1+z}{1+\bar z} & = \frac {1+\color{#3D99F6}{e^{i\theta}}}{1+\color{#3D99F6}{e^{-i\theta}}} & \small \color{#3D99F6}{\text{By Euler's formula,}} \\ & = \frac {1+\color{#3D99F6}{\cos \theta +i \sin \theta}}{1+\color{#3D99F6}{\cos \theta -i \sin \theta}} & \small \color{#3D99F6}{\text{Let }t=\tan \frac \theta 2, \ \cos \theta = \frac {1-t^2}{1+t^2}, \ \sin \theta = \frac {2t}{1+t^2}} \\ & = \frac {1+\color{#3D99F6}{\frac {1-t^2}{1+t^2} +i \frac {2t}{1+t^2}}}{1+\color{#3D99F6}{\frac {1-t^2}{1+t^2} -i \frac {2t}{1+t^2}}} \\ & = \frac {2+i2t}{2-i2t} \\ & = \frac {1+it}{1-it} \\ & = \frac {1+it}{1-it} \cdot \frac {1+it}{1+it} \\ & = \frac {1-t^2+i2t}{1+t^2} \\ & = \cos \theta + i \sin \theta \\ & = e^{i \theta} = z \end{aligned}$

Therefore, we have:

$\begin{aligned} \arg \left(\frac {1+z}{1+\bar z}\right) & = \arg (z) = \boxed{\theta} \end{aligned}$

I see no solution has a geometrical approach.. Will upload a geometrical Approach which is easier than the algebric one

Check z=1, θ=0. For this case (1+1)/(1+1)=1 and thus the argument is 0. However, that could mean the answer is θ or -θ. Check z=i, θ=π/2. In this case the argument of the argument function simplifies to simply i, which means the function has a value of π/2. This rules out the -θ case, meaning the answer must be θ.

The quantity asked in question is angle subtended by z and its conjugate at -1 which is clearly half of the angle subtended at the the origin ( which is 2 times theta).Hence the answer is theta.For verification we can check for z=i