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)

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

 

[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.