[tex]\text{Folosim urmatoare identitate:}\\
\text{Daca }z\in \mathbb{C\backslash R},\text{atunci: }z\cdot \overline{z}=|z|^2\\
\text{Avem ca:}\\
|z+w|=\sqrt 3|()^2\\
(z+w)(\overline{z}+\overline{w})=3\\
z\cdot \overline{z}+z\cdot \overline{w}+w\cdot \overline{z}+w\cdot \overline{w}=3\\
|z|^2+z\cdot \overline{w}+w\cdot \overline{z}+|w|^2=3\\
z\cdot \overline{w}+w\cdot \overline{z}+2=3\\
z\cdot \overline{w}+w\cdot \overline{z}=1(\text{relatia asta ne va ajuta mai tarziu})\\
\text{Notez }|z-w|=l\\
l^2=|z-w|^2\\
[/tex]
[tex]l^2=(z-w)(\overline{z}-\overline{w})\\
l^2=z\cdot \overline{z}-z\cdot \overline{w}-w\cdot \overline{z}+w\cdot \overline{w}\\
l^2=|z|^2+|w|^2-(z\cdot \overline{w}+w\cdot \overline{z})\\
l^2=1+1-1\\
l^2=1\Rightarrow l=1\\
\text{Deci }|z-w|=1[/tex]
