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$$\left|z_{1}-z_{2}\right|^{2}+\left|z_{1}+z_{2}\right|^{2}=2\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right)$$
for all $z_{1}, z_{2} \in \mathbb{C}$.
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Proof. We have
\begin{aligned} &\left|z_{1}-z_{2}\right|^{2}+\left|z_{1}+z_{2}\right|^{2} \\ =&\left(z_{1}-z_{2}\right) \overline{\left(z_{1}-z_{2}\right)}+\left(z_{1}+z_{2}\right) \overline{\left(z_{1}+z_{2}\right)} \\ =&\left(z_{1}-z_{2}\right)\left(\bar{z}_{1}-\bar{z}_{2}\right)+\left(z_{1}+z_{2}\right)\left(\bar{z}_{1}+\bar{z}_{2}\right) \\ =&\left(\left(z_{1} \bar{z}_{1}+z_{2} \bar{z}_{2}\right)-\left(z_{1} \bar{z}_{2}+z_{2} \bar{z}_{1}\right)\right)+\left(\left(z_{1} \bar{z}_{1}+z_{2} \bar{z}_{2}\right)+\left(z_{1} \bar{z}_{2}+z_{2} \bar{z}_{1}\right)\right) \\ =& 2\left(z_{1} \bar{z}_{1}+z_{2} \bar{z}_{2}\right)=2\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right) \end{aligned}
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