Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to show: $z_1+|a|^2z_2=z_2+|a|^2 z_1 \Rightarrow z_2=z_1$?

My proof: anthesis: Suppose $z_2\neq z_1$. Then $z_1+|a|^2z_2 \neq z_2+|a|^2 z_1$. RR So $z_2=z_1$.

share|improve this question
1  
If $|a|=1$ then $z_1+z_2=z_1+z_2$ does not imply $z_1=z_2$. –  njguliyev Sep 7 '13 at 10:38
    
The statement is false. Whenever $|a|^2=1$, the first statement trivially holds and the conclusion $z_2=z_1$ does not necessarily follow. –  Étienne Bézout Sep 7 '13 at 10:38
    
But if $|a| \ne 1$ then your "proof" is not correct. We have $(z_1-z_2)(|a|^2-1)=0$. –  njguliyev Sep 7 '13 at 10:39
    
@njguliyev: thanks. By the way $a\in \mathbb{D}$. I don't remember if $a=1 \in \mathbb{D}$? I guess this $1\in \mathbb{D}$. –  laovultai Sep 7 '13 at 10:53
1  
I guess $\mathbb{D}$ denotes the (open) unit disk, so if $a \in \mathbb{D}$, then $|a|^2-1 \neq 0$. –  Étienne Bézout Sep 7 '13 at 11:08

1 Answer 1

up vote 2 down vote accepted

If $a\in\{z\in\mathbb{C}:|z|<1\}$ then$|a|\ne 1$ and thus,

\begin{align} &z_1+|a|^2z_2=z_2+|a|^2 z_1 \\ &\iff z_1-z_2+|a|^2(z_2-z_1)=0 \\ &\iff (z_1-z_2)(1-|a|^2)=0 \\ &\iff z_1=z_2.\end{align}

share|improve this answer
    
I'd say, $a\in\{z\in\mathbb{C}:|z|\ne 1\}$. –  TZakrevskiy Sep 7 '13 at 11:55
    
@TZakrevskiy The OP mentioned in the comments that $a$ is a member of the open unit disk, so I used that set. But yes, $a\in\{z\in\mathbb{C}:|z|\ne 1\}$ covers all the cases where this statement is true. –  Alraxite Sep 7 '13 at 12:02

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.