Question

$z$ is a complex number. 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 $a\le b$

Given,
$a=|x|+|y|$
$b=\sqrt{2}|x+iy|=\sqrt{2}\sqrt{x^2+y^2}$
Let $z=rcos\theta +irsin\theta$
$a=rcos\theta +rsin\theta =r(cos\theta +sin\theta )$
Now, $cos\theta +sin\theta$ lies between $-\sqrt{2}$ to $\sqrt{2}$
So, $a_{max}=r\sqrt{2}$
Also, $b=\sqrt{2}|x+iy|=\sqrt{2}r$
Clearly $a_{max}=b$
$\Rightarrow a\le b$