Question

lf Z is a complex number and if $a=|x|+|y|$ and $b=\sqrt{2}|x+iy|$

then which of the following is true?

• A. $a\le b$
• B. $a>b$
• C. $a=b+\sqrt{2}$
• D. $a=b+2$

Answer 1

The correct option is A $a\le b$

$a=\left|x\right|+\left|y\right|$
$b=\sqrt{2}|x+iy|$
$Letx=rcos\theta y=rsin\theta$
$a=\left|rcos\theta \right|+\left|rsin\theta \right|$
$=\left|r\right|(\left|cos\theta \right|+\left|rsin\theta \right|)$
$⩽\sqrt{2}\left|r\right|=\sqrt{2}|x+iy|$
$\Rightarrow a⩽b$