Multiplicative invertive?

Question

Hi.
This is a follow up question to my question before. How does 1/1+i *1-i/1-i become 1-i/2?
Thank you.

Answer 1

The product of two conjugates is the difference of two squares. When complex numbers are involved, this becomes the sum of two squares.
If you put these expressions side by side, written out properly, multiply them out -carefully-, (FOIL the binomial expressions in the denominators), and then simplify, you will get the correct result. It's kind of hard to show computations like this on this limited text editor, but I'll do my best...
1/(1+i) * (1-i)/(1-i) =
(1*(1-i))/(1+i)(1-i) =
(1-i)/(1^2 + 1i - 1i - (i^2)) =
notice how the two middle terms cancel out in the denominator, +1i and -1i, also (i^2) = -1...
(1-i)/(1 - (-1)) =
notice how when you subtract a negative number, you chnage the sign on the subtrehend, and add...
(1-i)/(1+1) =
(1-i)/2
It's really cool how the product of two conjugate complex numbers yields the -sum- of two squares. I discovered this for myself when I was a kid, then when the math teacher said "you can't factor a sum of two squares" I blurted out what I'd discovered. I wasn't very popular as a kid.

Answer 2

1/(1 + i) x (1 - i)/(1 - i) = 1(1 - i)/(1 + i)(1 - i)

[remember (1 + i)(1 - i) = 1 - i^2 = 1 - (-1) = 2]

so it equals (1 - i)/2

Answer 3

just use FOIL for the denominator...

if you multiply it out...

(1/(1-i)) * (1-i)/(1-i)
= (1-i) / (1-i+i-1)
= (1-i) / 2

remember that i*i = -1 because i = root(-1)