Hi, how do I solve this "complex numbers" question?

Question

Let z=x+iy, where x, y are real numbers.
a) Find the real and imaginary parts of 1/(1-z), assuming z is not equals to 0.
b) If IzI=1 and z is not equals to 0, show that Re[1/(1-z)]=1/2.

Answer 1

1 / (1 - z) = 1 / [1 - (x + iy)] = 1 / [(1 - x) - iy], where (1-x) is the "real part" and y is the "imaginary part".

So to get it out of the denominator, multiply top and bottom by the "complex conjugate", which would be "(1-x) + iy":

1 / [(1 - x) - iy] * [(1-x) + iy]/[(1-x) + iy]
= [(1-x) + iy]/[(1-x)² - (iy)²]
= [(1-x) - iy]/[x² -2x + 1 - i²y²]
= [(1-x) - iy]/[x² -2x + 1 + y²]

So the real part would be:

(1-x)/[x² -2x + 1 + y²]

and the imaginary part would be

-y/[x² -2x + 1 - y²]

Very ugly.

For the second one, |z| = 1 means that x² + y² = 1.

So you can replace those 2 values in the denominator of the real part:

(1-x)/[x² + y² -2x + 1] = (1-x)/(1-2x+1)
= (1-x)/(2-2x)
= (1-x)/[2(1-x)]
=1/2

Since the (1-x)s cancel.

Answer 2

spoonish has the wrong numerator, but the right idea.

numerator is (1-x)-iy

if |z|=1, the x^2+y^2=1 or y^2=1-x^2. you need this to remove the y^2 in the denomiator. then just separate the real and imaginary parts.

Answer 3

1/(1-x-iy)
multiply by top and bottom by (1-x+iy)

(1+x+iy)
------------
(1-x)^2+y^2

multiply bottom out and separate numerator to real and imaginary

Answer 4

ask to google