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

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 $z_2=z_1$.

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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 $1\in \mathbb{D}$. – alvoutila Sep 7 '13 at 10:53
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

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