If $z_1, z_2$ are complex numbers and $u\in(0, \frac{\pi}{2})$ Prove that: $$\frac{|z_1|^2}{\cos^2u}+\frac{|z_2|^2}{\sin^2u}\ge|z_1|^2+|z_2|^2+2\text{Re}(z_1z_2)$$
I was just looking at the question above and do not know where to start. Had these been real numbers we were talking about, I would have immediately thought of using Andreescu as follows:
$$\frac{|z_1|^2}{\cos^2u}+\frac{|z_2|^2}{\sin^2u}\ge \frac{|z_1+z_2|^2}{\cos^2u+\sin^2u}=|z_1+z_2|^2$$
And this is where I get stuck. I assume that this is where imaginary numbers come into play. Could you please explain to me how to finish off this question and how to solve questions with complex numbers in general (e.g. reference to some source)?
Best Answer
Let $z_1=x_1+y_1i$ and $z_2=x_2+y_2i,$ where $x_1$, $x_2$, $y_1$ and $y_2$ are reals.
Thus, by C-S twice we obtain: $$\frac{|z_1|^2}{\cos^2u}+\frac{|z_2|^2}{\sin^2u}\geq\frac{(|z_1|+|z_2|)^2}{\cos^2u+\sin^2u}=|z_1|^2+|z_2|^2+2|z_1||z_2|=$$ $$=|z_1|^2+|z_2|^2+2\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}\geq|z_1|^2+|z_2|^2+2\sqrt{(x_1x_2-y_1y_2)^2}=$$ $$=|z_1|^2+|z_2|^2+2|Re(z_1z_2)|\geq|z_1|^2+|z_2|^2+2Re(z_1z_2).$$