If $z$ is a complex number of unit modulus and argument theta If $z$ is a complex number such that $|z|=1$ and $\text{arg} z=\theta$, then what is  $$\text{arg}\frac{1 + z}{1+ \overline{z}}?$$
 A: $$Arg\bigg(\frac{1+z}{1+\overline{z}}\bigg)=Arg\bigg(\frac{1+z}{1+(1/z)}\bigg)=Arg\bigg(\frac{1+z}{(z+1)/z}\bigg)=Arg\bigg(z\frac{1+z}{1+z}\bigg)=Arg(z)=\theta$$
A: Multiplying both numerator and denominator by $z$, we get:
$$\arg\left(\frac{1+z}{1+\bar{z}}\right)=\arg\left(\frac{z+z^{2}}{z+1}\right)=\arg\left(\frac{z(1+z)}{1+z}\right)=\arg\left(z\right)$$
We are told that $\arg(z)=\theta$, therefore:
$$\arg\left(\frac{1+z}{1+\bar{z}}\right)=\theta$$
A: Just a nice point in general, I found for you in my notes. If $z=\frac{x_1+iy_1}{x_2+iy_2}$, then you can try to write $z$ as $a+ib$ for some proper $a,b$. The way is a bit similar to @Mula's answer. We get, $$z=a+ib,~~ a=\frac{x_1x_2+y_1y_2}{x_2^2+y_2^2},~~ b=\frac{-x_1y_2+y_1x_2}{x_2^2+y_2^2}$$ Now, I think, you can find the answer by this another way also. Try it!
A: $$ \frac{1+z}{1 + \bar z} = \frac{1+z}{1 + \bar z} \times \frac{1+z}{1 + z} = \frac{(1+z)^2}{|1+z|^2} $$
the argument should be $ 2 \arg (1 +z) = 2 \arctan \left( y \over x+1\right)$
Let $x = \cos \theta$ and $y = \sin \theta $, we have $\arctan \left( \frac{y}{x+1}\right) = \arctan \left( \frac{\sin \theta}{\cos \theta+1}\right) = \arctan \left( \frac{\sin \frac{\theta}{2}}{\cos {\theta\over 2 }}\right)$
$2 \arg (1+z)^2 = \arg (z)$
A: $z$ can be written as  $\cos\theta+i\sin\theta\implies \bar z=\cos\theta-i\sin\theta$
So, $$\frac{1+z}{1+\bar z}=\frac{1+\cos\theta+i\sin\theta}{1+\cos\theta-i\sin\theta}$$
$$=\frac{2\cos^2\frac \theta2+2i\cos\frac\theta2\sin\frac\theta2}{2\cos^2\frac \theta2-2i\cos\frac\theta2\sin\frac\theta2}$$
$$=\frac{\cos\frac\theta2+i\sin\frac\theta2}{\cos\frac\theta2-i\sin\frac\theta2}$$
$$=\frac{\left(\cos\frac\theta2+i\sin\frac\theta2\right)^2}{\left(\cos\frac\theta2-i\sin\frac\theta2\right)\left(\cos\frac\theta2+i\sin\frac\theta2\right)}$$ assuming $\cos\frac\theta2\ne 0$ i.e., $\frac\theta2\ne (2n+1)\frac\pi2$ i.e., $\theta\ne(2n+1)\pi$ where $n$ is any integer
as $\theta=(2n+1)\pi, 1+\cos\theta\pm i\sin\theta=0$
$$\frac{1+z}{1+\bar z}=\cos\theta+i\sin\theta\text{ using de Moivre's formula} $$
$$\implies \frac{1+z}{1+\bar z}=z $$
$$\implies arg\left(\frac{1+z}{1+\bar z}\right)= arg(z) $$
