Hi guys I need help in proving this problem:
thanks guys. really need your immediate reply.
We have to prove that |(a-b)| = |(1-ab*)| , where |a| = 1
Hence, we need to show that the magnitude of the LHS = RHS
|a| = aa* = 1 => a = 1/a*
LHS = |(a-b)|
RHS = |(1-b*/a*)| => |(a*-b*)|/|a*| = |(a*-b*)|/1 = |(a*-b*)|
============> LHS = |(a-b)| = |(a*-b*)| = RHS ==================> It is PROVEN.
Since a & b are distinct complex numbers,
a = b + ci ; a* = b - ci ; b = x +yi ; b* = x - yi
LHS = |(a-b)| = |(b+ci-x-yi )| = |(b-x)+(c-y)i| = [(b-x)^2 + (c-y)^2]^(1/2)
RHS = |(a*-b*)| = |(b-ci-x+yi)| = |(b-x)+(y-c)i| = [(b-x)^2 + (y-c)^2]^(1/2)
Note: (c -y) = [-(c-y)]
let (c - y) be h,
h = -h => h^2 = (-h)^2 = |h|^2
let (b-x) be p,
LHS = (p^2 + h^2)^(1/2)
RHS = (p^2 + h^2)^(1/2) = LHS <=========== PROVEN
I don't think there's a mistake with the question, haha.
Btw, its not exactly a property, but just in case you are not aware.
First off, from the start, what we want to prove is this:
|(a-b)| = |1-ab*|
And, what you were asking is this: |(1-b/a)| = |(1-ab*)| (Why?)
Please note that the RHS of both equations are similar, which then tells me what you really were asking is how |(1-b/a)| come about from |(a-b)/(a)|.
What really happened happens within the absolute signs: (a-b)/(a) = ((a/a)-(b/a)) = (1-b/a).
I changed my workings which my teachers always complained messy; I hope you would now find it clearer, and easier to understand!
hehe.. i get it all actually except now for the first part..
why did you substitute |a|= 1 to WTF?
it's because WTF was the problem we wanted to prove so we can't assume that its equal to 1 unless we already proved it.
so i tried starting from the LHS of WTF and started manipulating it so that it became 1. what i got was
since 1 = |a|,
but b/a = ba' (where a' is the inverse of a) and since a' = a*/(|a|^2) = a*/(1^2) = a*,
b/a = ba*, therefore continuing from 1:
=|1-ba*|/|1-ab*| should be equal to 1.
which makes me suspect that our professor typed it wrong. hehe.