# Thread: A complex number problem

1. ## A complex number problem

Hi guys I need help in proving this problem:

2. ## i've got the 'answer'

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*

Hence,

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

Hence,

let (b-x) be p,

LHS = (p^2 + h^2)^(1/2)
RHS = (p^2 + h^2)^(1/2) = LHS <=========== PROVEN

3. Originally Posted by werepurple

=> |(a-b)| / |a| = |(a-b)/(a)| = |(1-b/a)| = |1-ab*|
why is |(1-b/a)| = |1-ab*|?
is there a property why this is true?

thanks for the rep btw.

4. hmmm.. you know what? I think the denominator should be 1-ba* not 1-ab*..

maybe our teacher just printed it wrong.. but I hope you do have a solution..

Thanks!!

5. 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!

6. 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

|(a-b)/(1-ab*)|=|a-b|/|1-ab*|

= (|a-b|/1)/|1-ab*|

since 1 = |a|,

=(|a-b|/|a|)/|1-ab*|

=|(a-b)/a|/|1-ab*|

=|1-(b/a)|/|1-ab*| (1)

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.

7. I made a mistake, I've corrected it. Lol, I was reading my workings, then I realized I made a really really stupid mistake. Haha, joined this forum just yesterday, not really careful doing math using my keyboard lols.

8. hey thanks!!
actually I finally realized the missing step earlier in class...hehe.. I realized that

|1-ba*| = |(1-ba*)*|
= |1*-(ba*)*|
= |1-ab*|

anyways, you also proved it another way so thanks!!!