Question

If $z$ is a complex number such that ${\mathop{\rm Re}\nolimits} (z) = {\mathop{\rm Im}\nolimits} (z)$, then
(A) ${\mathop{\rm Re}\nolimits} ({z^2}) = 0$ (B) ${\mathop{\rm Im}\nolimits} ({z^2}) = 0$ (C) ${\mathop{\rm Re}\nolimits} ({z^2}) = {\mathop{\rm Im}\nolimits} ({z^2})$ (D) ${\mathop{\rm Re}\nolimits} ({z^2}) = - {\mathop{\rm Im}\nolimits} ({z^2})$

Answer 1

Hint- Assume $z = x + iy$ as a complex number, where $x$ and $y$ are real numbers. To solve this question we need to find the real part and imaginary part of ${z^2}$.

Complete Step by step solution:
Let $z = x + iy$, where $x \ne 0 $and $y \ne 0$
Also,
$x$ is real part of $z$ i.e. ${\mathop{\rm Re}\nolimits} (z)$ and,
$y$ is imaginary part i.e. ${\mathop{\rm Im}\nolimits} (z)$
As given in the question,
${\mathop{\rm Re}\nolimits} (z) = {\mathop{\rm Im}\nolimits} (z)$
$ \Rightarrow x = y$ equation (1)
Now, we take square of $z$,
${z^2} = {(x + iy)^2}$
${z^2} = {x^2} - {y^2} + 2ixy$
Here again we have real part and imaginary part as ${\mathop{\rm Re}\nolimits} ({z^2})$ and ${\mathop{\rm Im}\nolimits} ({z^2})$.
 For ${\mathop{\rm Re}\nolimits} ({z^2})$,
${\mathop{\rm Re}\nolimits} ({z^2}) = {x^2} - {y^2}$
$ \Rightarrow {\mathop{\rm Re}\nolimits} ({z^2}) = {x^2} - {x^2}$ [from equation (1)]
$ \Rightarrow {\mathop{\rm Re}\nolimits} ({z^2}) = 0$ equation (2)
And for${\mathop{\rm Im}\nolimits} ({z^2})$,
${\mathop{\rm Im}\nolimits} ({z^2}) = 2xy$
Which can’t be $0$ as $x \ne 0 $ and $y \ne 0$
So ${\mathop{\rm Im}\nolimits} ({z^2}) \ne 0$ equation (3)
From equation (2) and equation (3) we have some conclusions as
${\mathop{\rm Re}\nolimits} ({z^2}) = 0$, ${\mathop{\rm Im}\nolimits} ({z^2}) \ne 0$ and therefore ${\mathop{\rm Re}\nolimits} ({z^2}) \ne {\mathop{\rm Im}\nolimits} ({z^2})$
Clearly, option (A) is the only correct option.


Note: $i = \sqrt { - 1} $ and ${i^2} = - 1$.
The real number $a$ is called the real part of the complex number $a + ib$; the real number $b$is called its imaginary part. To emphasize, the imaginary part does not include a factor $i$; that is, the imaginary part is $b$, not $ib$.