Question

If $z=x+iy,x,y$ real, then $\left|x\right|+\left|y\right|\le k\left|z\right|$, where $k$ is equal to

• A. $1$
• B. $\sqrt{2}$
• C. $\sqrt{3}$
• D. None of these

Answer 1

Answer: (B)

$\sqrt{2}$

For every $a\in R,\left|a\right|=\sqrt{a^2}\Rightarrow {\left|a\right|}^2=a^2$

Now, ${(\left|x\right|-\left|y\right|)}^2\ge 0\Rightarrow {\left|x\right|}^2+{\left|y\right|}^2-2\left|x\right|\left|y\right|\ge 0$

$\Rightarrow 2\left|x\right|\left|y\right|\le {\left|x\right|}^2+{\left|y\right|}^2$

$\Rightarrow {\left|x\right|}^2+{\left|y\right|}^2+2\left|x\right|\left|y\right|\le {2\left|x\right|}^2+2{\left|y\right|}^2$

$\Rightarrow {(\left|x\right|+\left|y\right|)}^2\le 2(x^2+y^2)\Rightarrow {(\left|x\right|+\left|y\right|)}^2\le 2{\left|z\right|}^2$

$\therefore \left|x\right|+\left|y\right|\le \sqrt{2}z$