Question

Prove that for any complex number $z$,
$\left|Re\left(z\right)\right|+\left|Im\left(z\right)\right|\le \left|z\right|\sqrt{2}$
or $\left|x\right|+\left|y\right|\le \sqrt{2}|x+iy|$

Answer 1

Let $z=x+iy={re}^{i\theta }=r(\cos \theta +i\sin \theta )\left|x\right|+\left|y\right|$
$=r[\left|\cos \theta \right|+\left|\sin \theta \right|]$
Squaring both sides $(r+ivealways)$
${[\left|x\right|+\left|y\right|]}^2=r^2[1+\left|2\sin \theta \cos \theta \right|]$
$=r^2[1+\left|\sin 2\theta \right|]$
$<r^2[1+1]\because \left|\sin x\right|\le 1$
$\therefore \left|x\right|+\left|y\right|\le r\sqrt{2}$
$\therefore \left|Rez\right|+\left|Imz\right|\le \sqrt{2}\left|z\right|\because \left|z\right|=r$
or $\therefore \left|x\right|+\left|y\right|\le \sqrt{2}|x+iy|$