Question

z is a complex number. If $a=\left|x\right|+\left|y\right|$ and $b=\sqrt{2}|x+iy|$

Answer 1

Now,

$\begin{array}{c}a=\left|x\right|+\left|y\right|b=\sqrt{2}\left|x+iy\right|\\ a^2={\left|x\right|}^2+{\left|y\right|}^2+2\left|x\right|\left|y\right|\\ b^2=2\left(x^2+y^2\right)\\ \therefore a^2=x^2+y^2+2\left|x\right|\left|y\right|\\ \because {\left(\left|x\right|-\left|y\right|\right)}^2={\left|x\right|}^2+{\left|y\right|}^2-2\left|x\right|\left|y\right|\\ {\left|x\right|}^2+{\left|y\right|}^2\ge 2\left|x\right|\left|y\right|\\ x^2+y^2\ge 2\left|x\right|\left|y\right|\\ b^2\ge a^2\\ b\ge a\end{array}$