Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is Exercise EP $14$ from Fernandez and Bernardes's book Introdução às Funções de uma Variável Complexa (in Portuguese). The authors ask us to prove that the inequality $$\sqrt{2}|z|\geq |\operatorname{Re}z|+|\operatorname{Im}z|$$ holds for all $z\in \mathbb{C}.$

I know that $|z|\geq\max\{|\operatorname{Re}z|,|\operatorname{Im}z|\}$, but all I've managed to do is show the obvious: $2|z|\geq |\operatorname{Re}z|+|\operatorname{Im}z|.$

I would appreciate a hint here.

share|improve this question
    
Please don't use display math ($$...$$) in titles. :) –  cardinal Mar 4 '12 at 21:56
1  
@cardinal: I won't do it anymore! Thanks for helping me to do the right thing! –  spohreis Mar 4 '12 at 22:17
1  
It happens pretty often. It's not a big deal, so I went ahead and edited it. Cheers. :) –  cardinal Mar 4 '12 at 22:20

1 Answer 1

up vote 6 down vote accepted

Let $z=x+iy$, i.e. $x=\operatorname{Re}z$ and $y=\operatorname{Im}z$. Then $$(\sqrt{2}|z|)^2=2|z|^2=2x^2+2y^2.$$ On the other hand, $$(|\operatorname{Re}z|+|\operatorname{Im}z|)^2=(|x|+|y|)^2=x^2+y^2+2|xy|.$$ This implies that $$(\sqrt{2}|z|)^2-(|\operatorname{Re}z|+|\operatorname{Im}z|)^2=x^2+y^2-2|xy|=(|x|-|y|)^2\geq 0$$ which implies that $$\sqrt{2}|z|\geq |\operatorname{Re}z|+|\operatorname{Im}z|$$ as required.

Remarks: This also shows that the equality holds if and only if $|\operatorname{Re}z|=|\operatorname{Im}z|$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.