Question

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

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

Answer 1

The correct option is A [0, 4]

$z=x+iy$
And $|z{|}^2=16$
$|z|=4$
If z is purely real, $x=\pm 4$ and $||x|-|y||=4$ ...(i)
If z is purely imaginary $y=\pm 4i$ and $||x|-|y||=4$ ...(ii)
Now if $|x|=|y|$
Thus
$2x^2=2y^2=16$
$x=y=\pm 2\sqrt{2}$
Under such case
$||x|-|y||=0$
Hence the range of values of $||x|-|y||$ is $[0,4]$