How to show that $|z_1+z_2|+|z_1-z_2| = |z_1+\sqrt{z^2_1-z^2_2} |+|z_1-\sqrt{z^2_1-z^2_2} | $?

Question

How to show that $$|z_1+z_2|+|z_1-z_2| = |z_1+\sqrt{z^2_1-z^2_2} |+|z_1-\sqrt{z^2_1-z^2_2} | $$

Is this question correct?

I tried my best to prove it

But got nowhere near, Any help

Answer 1

Let $a=z_1+z_2$ and $b=z_1-z_2$, then we want to show that $$|a|+|b|=\left|\frac{a+b}{2}+\sqrt{ab}\right|+\left|\frac{a+b}{2}-\sqrt{ab}\right|=\frac{1}{2}\left|\sqrt{a}+\sqrt{b}\right|^2+\frac{1}{2}\left|\sqrt{a}-\sqrt{b}\right|^2$$ Now we know that \begin{align*} |z+w|^2 & =(z+w)(\bar{z}+\bar{w})=|z|^2+2(z\bar{w}+w\bar{z})+|w|^2\\ |z-w|^2 & =(z-w)(\bar{z}-\bar{w})=|z|^2-2(z\bar{w}+w\bar{z})+|w|^2 \end{align*} Let $z=\sqrt{a}$ and $w=\sqrt{b}$. Recall that $\left| \sqrt{a}\right|^2=\left|(\sqrt{a})^2\right|=|a|$. Hopefully you can take it from here.