If $ \vert z \vert = a $ then what is the real part of $ ( \frac{z-a}{z+a} ) $ I am beginner to complex numbers.
I am having no idea to solve this:
Let $ z $ be any non - zero complex number.
If $ \vert z \vert = a $ , then what is the real part of $ ( \frac{z-a}{z+a} ) $ ?

"a" is greater than 0.
Use $Re(w) = \frac {w + \bar{w}} 2, w \in \mathbb{C}$ for $w=\frac{z-a}{z+a}$ then make the calculus.

You'll get something like: $\frac 1 2 (\frac{z-a}{z+a} + \frac{\bar{z}-a}{\bar{z}+a}) = \frac 1 2 (\frac {a^2 + za-a\bar{z} - a^2 + a^2 - za+a\bar{z} - a^2}{(z+a)(\bar{z} + a)}) = 0$
WLOG $z=ae^{2it}$ where $t$ is real

$$\dfrac{z-a}{z+a}=\dfrac{e^{2it}-1}{e^{2it}+1}=\dfrac{e^{it}-e^{-it}}{e^{it}+e^{-it}}=\dfrac{2i\sin t}{\cos t}=?$$