What is the real part of this complex function?

Question

F(z) = (1 + Z) / (1 - Z), where Z = x + iy

Answer 1

1+Z = x+1+iy
1-Z = 1-x-iy

(x+1+iy) / (1-x-iy)
= (x+1+iy)(1-x+iy) / ((1-x)² + y²)
= (x - x² + ixy + 1 - x + iy + iy - ixy - y²) / (1 - 2x + x² + y²)

Re(1 + Z) / (1 - Z)
= (x - x² + 1 - x - y²) / (1 - 2x + x² + y²)
= (1 - x² - y²) / (1 - 2x + x² + y²)

Answer 2

f(x+iy) = (1+x + iy) / (1-x - iy)

Multiply both the numerator and denominator by the conjugate:

(1+x + iy)(1-x + iy) / ((1-x)² + y²)
(1-x² - y² + i(1+x)y + i(1-x)) / ((1-x)² + y²)
(1-x²-y²) / ((1-x)²+y²) + i 2y / ((1-x)² + y²)

So the real part of f is (1-x²-y²) / ((1-x)²+y²).

Answer 3

F(z) = (1+x+iy)/(1-x-iy)
convert to polar form and divide
F= R1 angle(q1) / R2 angle(q2)
R1=sqrt{(1+x)^2+y^2}
angle q1=tan-1(y/1+x)
R2=sqrt{(1-x)^2+y^2}
angle q2=tan-1(-y/(1-x))
F=S angle (T)
S=R1/R2
angle T= angle (q1 -q2)
Real(F) = S Cos(T)