# Intriguing question on Complex Numbers

• Jan 10th 2010, 11:06 PM
bandedkrait
Intriguing question on Complex Numbers
I found a simple but interesting problem which can be solved using highschool concepts regarding complex numbers... (I posted it here since it's purely optional, those interested may try.)

Consider three real numbers $\displaystyle x,y,z$ none equal to zero.

$\displaystyle \alpha,\beta,\gamma$ are three complex numbers such that

http://latex.codecogs.com/gif.latex?...mma \right |=1

if $\displaystyle x+y+z=0$ , and http://latex.codecogs.com/gif.latex?...a y+\gamma z=0,

Prove that : http://latex.codecogs.com/gif.latex?...=\beta =\gamma
• Jan 18th 2010, 10:02 AM
Dinkydoe
Given $\displaystyle \alpha,\beta,\gamma$ on the complex-unit ring in $\displaystyle \mathbb{C}$

First we oberve that: $\displaystyle x+y+z = 0 \Leftrightarrow x = -(y+z)$

Hence:
$\displaystyle \alpha x+\beta y +\gamma z = 0\Leftrightarrow$
$\displaystyle -\alpha(y+z)+\beta y + \gamma z = 0 \Leftrightarrow$
$\displaystyle \beta y + \gamma z = \alpha y + \alpha z \Leftrightarrow$
$\displaystyle y+[\beta^{-1}\gamma]z = [\beta^{-1}\alpha](y+z)$

Since $\displaystyle |\beta^{-1}\alpha| = 1$ we obtain:

$\displaystyle |y+[\beta^{-1}\gamma]z|= |y+z| \Rightarrow \beta^{-1}\gamma = 1$. Thus $\displaystyle \beta = \gamma$. And from this follows $\displaystyle \alpha=\beta=\gamma.$