# Proving an identity relating to the complex modulus: $z\bar{a}+\bar{z}a \leq 2|a||z|$

Show that $z\bar{a}+\bar{z}a \leq 2|a||z|$

I was able to check this using some examples which is not ideal for a mathematician way of proving a problem/case/theorem. I couldn't generalize it (prove it). I will be glad if I can be given the hint/proof for this problem. Thank you.

• Jun 6, 2017 at 8:19

First, notice that one has: $$z\overline{a}+\overline{z}a=z\overline{a}+\overline{z\overline{a}}=2\textrm{Re}(z\overline{a}).$$ Now, recall that: $$\forall w\in\mathbb{C},\textrm{Re}(w)\leqslant |w|.$$ Whence the result.

First you have to convince yourself that the lhs of your inequality is a real. Or else, the exercise has no sense.

It is the case because $\overline{z\overline a+\overline za} = \overline za + z\overline a$, so your number is real.

Now all you need is the triangle inequality : $$z\overline a+\overline za \le \left|z\overline a+\overline za\right| \le \left|z\overline a\right| + \left|\overline za\right| = \left|\overline z\right|\left|a\right|+\left|z\right|\left|\overline a\right| = 2\left|z\right|\left|a\right|$$

Hints:

By a formula for the complex norm $$0\leq|z+a|^2=(z+a)\overline{(z+a)}=(z+a)(\overline{z}+\overline{a})=|z|^2+z\overline{a}+\overline{z}a+|a|^2$$

By the triangle inequality: $$|z+a|^2\leq(|z|+|a|)^2=|z|^2+2|z||a|+|a|^2$$

I want to add another point of view to this question although it has already been answered. An identity which can be quite helpful for your intuition is the following*:

$$\bar z_1 z_2 = (\vec z_1\cdot \vec z_2)+i\hat k\cdot (\vec z_1\times \vec z_2).$$

This identity can halp you in lots of questions where one complex number multiplied by the conjugate of another number appears.

Here the dot and cross on the LHS coresspond to the dot and cross vector products when the complex numbers a and b are treated as vectors (i.e. $(a+ib)\leftrightarrow(a\hat\imath+b\hat \jmath)$).

Hence by Cauchy-Schwarz inequality,

$$z\bar{a}+\bar{z}a = 2(\vec z\cdot \vec a)\leq 2|z||a|.$$

(the asymmetry of the cross product causes it to be cancelled).

Look! I solved it in one line! :)

*(Since all complex numbers lie in one plane, the cross product reduces to a vector in the $\hat k$ direction and we want only a number not a vector so I take the k component).

• You can avoid the "neglect $\hat k$" part by just writing $i\hat k\cdot(\vec z_1\times \vec z_2)$. Jun 6, 2017 at 8:39
• @celtschk Hmm you are right .... Jun 6, 2017 at 11:23