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Let us consider $z_1, z_2\in \mathbb C$; we have:

$$\bar z_1 z_2+z_1 \bar z_2=2\Re(\bar z_1 z_2)$$

it is easy to prove if we put $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$. But suppose we do not want to use $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$. How is it possible to prove the following relationship?

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    $\begingroup$ $z+\bar{z}=2Re(z)$ $\endgroup$ – Semsem Sep 29 '14 at 21:41
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Hint: $z_1\bar{z}_2 = \overline{\bar{z}_1z_2}$.

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