# prove $2Re (z_1 \bar{z_2}) \leq 2 |z_1||z_2|$

prove that $\forall z_1,z_2 \in \mathbb{C}$ $$2Re (z_1 \bar{z_2}) \leq 2 |z_1||z_2|$$

Attempt

$Re(z_1 \bar{z_2})=x_1x_2+y_1y_2$ and $|z_1||z_2|=\sqrt{x_1^2+y_1^2}\sqrt{x_2^2+y_2^2}$ so \begin{aligned} 2Re (z_1 \bar{z_2}) & \leq 2 |z_1||z_2| && \text{div by 2} \\ x_1x_2+y_1y_2 &\leq \sqrt{x_1^2+y_1^2}\sqrt{x_2^2+y_2^2} \\ (x_1x_2+y_1y_2)^2 &\leq (\sqrt{x_1^2+y_1^2}\sqrt{x_2^2+y_2^2})^2 && \text{square rooting} \\ x_1^2*x_2^2+2x_1 x_2 y_1 y_2 +y_1^2 y_2^2 &\leq x_1^2 x_2^2 +x_1 ^2 y_2^2 +y_1^2 x_2^2 +y_1^2 y_2 ^2 \\2x_1 x_2 y_1 y_2 &\leq x_1^2 y_2^2 +y_1^2 x_2^2& \\ 0&\leq x_1^2 y_2^2 +y_1^2 x_2^2 -2x_1x_2 y_1 y_2 \\ 0 &\leq (\text{stuck})^2 \end{aligned} Not sure where to go from there. I want to say i have done this in the past but have a lapse. Also i am studying with someone else and is saying $Re(z_1 \bar{z_2}) \leq |z_1z_2|=|z_1||z_2|$ which would be simpler.

• For a different proof, try writing the variables in polar form i.e. $z_i = |z_i|e^{i\theta_i}$ for $i=1,2$. – Semiclassical Aug 26 '16 at 20:23

Since you proved that the inequality is equivalent to $(\text{stuck})^2\ge0$, and the latter is true because $\text{stuck}=x_1y_2 - x_2y_1 \in \mathbb R$, you are done. But there's a small issue that you need to take care of: LHS in the second inequality may be negative, so work with absolute values from the beginning
Another way is prove that for any $z$, $|\Re z| \le |z|$ (it's pretty straightforward), so $\Re(z_1 \bar z_2) \le | z_1\bar z_2| = |z_1||z_2|$
Your inequality is equivalent to $$[\Re(z_1\bar z_2)]^2\le|z_1|^2|z_2|^2$$ Observe now that $|z_2|=|\overline {z_2}|$, so the inequality becomes $$[\Re(z_1\bar z_2)]^2\le|z_1\overline {z_2}|^2$$
but putting $w=a+ib$ an arbitrary complex number, you always have $$|w|^2=a^2+b^2\ge a^2$$ which is exactly your inequality with $w=z_1\overline{z_2}$.