Prove Complex Inequality |Re(z)| + |Im(z)| \le \sqrt{2}|z|

[cscott]

Homework Statement

Prove

$$|Re(z)| + |Im(z)| \le \sqrt{2}|z|$$

The Attempt at a Solution

I was going to try to get from the LHS to RHS

I can only see squaring then square-rooting the LHS and somehow getting to

something involving $$\sqrt{|Re(z)|^2 + |Im(z)|^2}$$ to recover |z|, but I don't see it... any hints?

 

[rochfor1]

Can you find an equality relating just |Re(z)| and |z|?

 

[cscott]

Mmm..

$$z + \bar{z} = 2 Re(z)$$

[tex]|z +\bar{z}| = 2 |Re(z)|[/itex]<br /> <br /> $$|z| + |\bar{z}| \ge 2 |Re(z)|$$<br /> <br /> How about that?[/tex]

 

[qntty]

re(z) = |z|cos(x)
im(z) = |z|sin(x)

use trig identities

 

[rochfor1]

cscott, your last inequality is almost the right one, in fact, you can get the one I'm thinking of from it. You're making it a bit more complicated than it needs to be though.

 

[cscott]

How about,

$$Re(z) + Im(z) = |z| \left (\cos \theta + \sin \theta \right )$$

cos+sin has a maximum absolute value of $\sqrt{2}$ so,

$$|Re(z)| + |Im(z)| \le \sqrt{2}|z|$$

I know I'm still making this more difficult than necessary :s

 

[jgens]

That works. If $z = a + bi$ then you can also prove the inequality from the fact that $(a-b)^2 \geq 0$.

 

[rochfor1]

It definitely works. The inequality I was looking for was |Re(z)|<=|z|, by the way.