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How can I show this inequality $\sqrt{2}|z|\geq |\mathrm{Re} (z)|+|\mathrm{Im}(z)| $ please give me some hint. Which result is useful to show this. please help me out.thanks in advance.

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up vote 3 down vote accepted

You need to think about the geometric representation of complex numbers. Given a rectangle with sides $a = |\Im(z)|$ and $b = |\Re(z)|$, show that its diagonal is not shorter than $\frac{a+b}{\sqrt 2}$.

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Just write down $z$ as $a+bi$ with $a,b\in \mathbf R$. You can assume without loss of generality that $a,b\geq 0$. Then it's pretty much pure (elementary) algebra.

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If $z=x+iy$ where x,y are real, $\sqrt2|z|=\sqrt{2(x^2+y^2)}$ which will be ≥ |x|+|y| as Re(z)=x and Im(z)=y,

iff $2(x^2+y^2)≥x^2+y^2+2|x||y|$ (as $|x|^2=x^2$ for real x)

iff $(|x|-|y|)^2≥0$ which is true as x,y are real

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