Question

lf $Z=x+iy$ is a complex number then $|x|+|y|\le$ ?

• A. $\sqrt{2}|z|$
• B. $2|Z|$
• C. $\frac{\sqrt{2}}{Z}$
• D. $|Z|$

Answer 1

The correct option is A $\sqrt{2}|z|$

$z=x+iy=r\cos \theta +ir\sin \theta where\left|z\right|=r$
Now, $\left|x\right|+\left|y\right|=\left|r\cos \theta \right|+\left|r\sin \theta \right|$
$=|r|\{\cos \theta |+|\sin \theta |\}$
But $-\sqrt{2}<\cos s\theta +sin\theta <\sqrt{2}$
$\Rightarrow |x|+|y|⩽\sqrt{2}|r|$
$⩽\sqrt{2}|z|$