Question

If $z=x+iy$ and $x^2+y^2=16$, then the range of $||x|-|y||$ is

• A. $[0,4]$
• B. $[0,2]$
• C. $[2,4]$
• D. None of these

Answer 1

The correct option is A $[0,4]$

Here, $x=\cos \theta ,y=\sin \theta$
$||x|-|y||=|4|\cos \theta |-4|\sin \theta ||$
$=4||\cos \theta |-|\sin \theta ||$
$=4\sqrt{1-2|\cos \theta ||\sin \theta |}$
$=4\sqrt{1-|\sin 2\theta |}$
Hence, the range is $[0,4]$