I'm given the following problem:
Problem. Let $w$ and $z$ be complex numbers. Show that $|w\overline{z}+z\overline{w}| \leq 2|wz|$. You must use direct proof.
Attempt. Let $w = a+bi$ and $z = c+di$, for $a,b,c,d\in\mathbb{R}$.
Then $|w\overline{z} + z\overline{w}| = |(a+bi)(c-di) + (a-bi)(c+di)| = |2(ac+bd)| = 2\sqrt{(ac+bd)^2} = 2(ac+bd)$.
$2|wz| = 2\sqrt{(a+b)^2 + (c+d)^2} = 2\sqrt{2(ab+cd) + c^2 + d^2 + a^2 + b^2}$
…I'm not sure where to proceed from here.
Alternatively, we could express $w$ and $z$ is "polar exponential" form, letting $w = r_1e^{i\theta_1}$ and $z =r_2e^{i\theta_2}$. Then,
$|w\overline{z} + z\overline{w}| = |r_1e^{i\theta_1}r_2e^{-i\theta)2} + r_2e^{i\theta_2}r_1e^{i\theta_1}| = \frac{r_1}{r_2}+\frac{r_2}{r_1} = \frac{r_1^2 + r_2^2}{r_1r_2}$.
Unfortunatey, it isn't clear that $2r_1r_2 \geq \frac{r_1^2 + r_2^2}{r_1r_2}$ here, either.
How may I proceed? Is this an effective approach?
Best Answer
Remember that $|\overline{a}| = |a|$ and $|ab| = |a||b|$.
Based on such results as well as on the triangle inequality, we get that
\begin{align*} |w\overline{z} + \overline{w}z| & \leq |w\overline{z}| + |\overline{w}z|\\\\ & = |w||\overline{z}| + |\overline{w}||z|\\\\ & = |w||z| + |w||z|\\\\ & = |wz| + |wz|\\\\ & = 2|wz| \end{align*}
Hopefully this helps !