Suppose w=(z+1)/(z-1) where z=a+bi, a, b in RR. If |z|=1, can you find Re(w)?

i.e. can you find the real part of $w$?

Nov 22, 2016

$R e \left[w\right] = 0$

Explanation:

Given a complex number $w$ then

$R e \left[w\right] = \frac{w + \overline{w}}{2}$ so

$w + \overline{w} = \frac{z + 1}{z - 1} + \frac{\overline{z} + 1}{\overline{z} - 1} =$
$= \frac{\left(z + 1\right) \left(\overline{z} - 1\right) + \left(z - 1\right) \left(\overline{z} + 1\right)}{\left(z - 1\right) \left(\overline{z} - 1\right)} =$
$2 \frac{z \overline{z} - 1}{z \overline{z} - z - \overline{z} + 1} = 2 \frac{{a}^{2} + {b}^{2} - 1}{{a}^{2} + {b}^{2} - 2 a + 1}$

and

$R e \left[w\right] = \frac{{a}^{2} + {b}^{2} - 1}{{a}^{2} + {b}^{2} - 2 a + 1}$

but ${a}^{2} + {b}^{2} = 1$ so

$R e \left[w\right] = 0$

NOTE:

If $w = \frac{a + i b + 1}{a + i b - 1} = \frac{\left(a + i b + 1\right) \left(a - i b - 1\right)}{\left(a + i b - 1\right) \left(a - i b - 1\right)} = \frac{{a}^{2} + {b}^{2} - 1 - 2 i b}{{\left(a - 1\right)}^{2} + {b}^{2}}$

analogously

$\overline{w} = \frac{a - i b + 1}{a - i b - 1} = \frac{{a}^{2} + {b}^{2} - 1 + 2 i b}{{\left(a - 1\right)}^{2} + {b}^{2}}$