If $z$ is a complex number such that Re $\left( z \right) = {\rm{Im}}\;\left( z \right)$, then

${\mathop{\rm Re}\nolimits} \left( {{z^2}} \right) = 0$

${\mathop{\rm Im}\nolimits} \left( {{z^2}} \right) = 0$

${\mathop{\rm Re}\nolimits} \left( {{z^2}} \right) = {\rm{Im}}\;\left( {{z^2}} \right)$

${\mathop{\rm Re}\nolimits} \left( {{z^2}} \right) = - {\rm{Im\;}}\left( {{{\rm{z}}^2}} \right)$

JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

1 of 1

Application +4 | -1

z

is a complex number such that Re

\left( z \right) = \mathrm{Im}\;\left( z \right)

, then

View answer and explanation

Correct option: A

{\mathop\mathrm{Re}\nolimits} \left( {{z^2}} \right) = 0

Option A (

{\mathop\mathrm{Re}\nolimits} \left( {{z^2}} \right) = 0

) is correct because if the real and imaginary parts of z are equal, then the real part of z² is zero.