Question

If $z$ is a complex number satisfying $|z{|}^2–|z|-2<0$, then the value of $|z^2+zsin\theta |$, for all values of $\theta$, is

• A. equal to 4
• B. equal to 6
• C. more than 6
• D. less than 6

Answer 1

The correct option is D

less than 6

$|z{|}^2-|z|-2<0$

$\Rightarrow (|z|-2)(|z|+1)<0$

$\Rightarrow |z|<2$

Now $|z^2+zsin\theta |\le |z{|}^2+|zsin\theta |\le |z{|}^2+|z|<4+2=6$