1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ For a different proof, try writing the variables in polar form i.e. $z_i = |z_i|e^{i\theta_i}$ for $i=1,2$. $\endgroup$ – Semiclassical Aug 26 '16 at 20:23
6
$\begingroup$

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 $\Re(z_1 \bar z_2) \le | z_1\bar z_2| = |z_1||z_2|$

$\endgroup$
2
$\begingroup$

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 $w=z_1\overline{z_2}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.