Q. Re $\left( \mathcal{z}^{2} \right) = 1$ is represented by
A) The circle $x^{2} + y^{2} = 1$
B) The hyperbola $x^{2} - y^{2} = 1$
C) Parabola or a circle
D) All of the above
Explanation
(b)
Let $\mathcal{z =}x + iy$
$\Rightarrow \ \mathcal{z}^{2} = x^{2} - y^{2} + 2i\ xy$
$\Rightarrow Re\left( \mathcal{z}^{2} \right) = Re(x^{2} - y^{2} + 2ixy)$
$\Rightarrow 1 = x^{2} - y^{2}\ \ \ \ \ \ \ \ \ \ \lbrack\because Re\left( \mathcal{z}^{2} \right) = 1$ (given)]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.