Complex inequality

  • Thread starter cscott
  • Start date
  • #1
782
1

Homework Statement



Prove

[tex]|Re(z)| + |Im(z)| \le \sqrt{2}|z|[/tex]

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 [tex]\sqrt{|Re(z)|^2 + |Im(z)|^2}[/tex] to recover |z|, but I don't see it... any hints?
 

Answers and Replies

  • #2
256
0
Can you find an equality relating just |Re(z)| and |z|?
 
  • #3
782
1
Mmm..

[tex]z + \bar{z} = 2 Re(z)[/tex]

[tex]|z +\bar{z}| = 2 |Re(z)|[/itex]

[tex]|z| + |\bar{z}| \ge 2 |Re(z)|[/tex]

How about that?
 
  • #4
290
2
re(z) = |z|cos(x)
im(z) = |z|sin(x)

use trig identities
 
  • #5
256
0
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.
 
  • #6
782
1
How about,

[tex]Re(z) + Im(z) = |z| \left (\cos \theta + \sin \theta \right )[/tex]

cos+sin has a maximum absolute value of [itex]\sqrt{2}[/itex] so,

[tex]|Re(z)| + |Im(z)| \le \sqrt{2}|z|[/tex]

I know I'm still making this more difficult than necessary :s
 
  • #7
jgens
Gold Member
1,581
50
That works. If [itex]z = a + bi[/itex] then you can also prove the inequality from the fact that [itex](a-b)^2 \geq 0[/itex].
 
  • #8
256
0
It definitely works. The inequality I was looking for was |Re(z)|<=|z|, by the way.
 

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