Question

If $z$ is the complex number such that $\operatorname{Re} (z) = \operatorname{Im} (z)$ then
A. $\operatorname{Re} ({z^2}) = 0$
B. $\operatorname{Im} ({z^2}) = 0$
C. $\operatorname{Re} ({z^2}) = \operatorname{Im} ({z^2})$
D. $\operatorname{Re} ({z^2}) = - \operatorname{Im} ({z^2})$

Answer 1

Hint:
Let $z = a + ib$ so $\operatorname{Re} (z) = a,\operatorname{Im} (z) = b$ and it is also given that $a = b$ so $z = a + ia = a(1 + i)$ and now we can find ${z^2}$.

Complete step by step solution:
Here we are given that $z$ is the complex number and we know that complex number is formed by the real part denoted by $\operatorname{Re} (z)$ and the imaginary part which is given by $\operatorname{Im} (z)$ and here real part is the constant term while the imaginary part is the coefficient of $i$ which is iota and is given as $i = \sqrt { - 1} $. So let us assume $z = a + ib$
So here $a$ is the real part and $b$ is the imaginary part which means that $\operatorname{Re} (z) = a,\operatorname{Im} (z) = b$
And we are also given that imaginary and the real part of the complex function are equal which means that $a = b$
So as we know that $a = b$ so we can replace $b$ with $a$
So
$z = a + ib$
$z = a + ia = a(1 + i)$
Now we need to find the ${z^2}$
So upon squaring both the sides we will get that
${z^2} = {a^2}{(1 + i)^2}$
Here we also know that ${(a + b)^2} = {a^2} + {b^2} + 2ab$
So applying this formula in the above equation we will get that
${z^2} = {a^2}(1 + {i^2} + 2i)$
$ = {a^2}(1 + ( - 1) + 2i)$
$ = {a^2}(1 - 1 + 2i)$
$ = 2{a^2}i$
So we know that ${z^2}$ is another complex number whose real part is $0$ and imaginary part is $2{a^2}$

So we can say that $\operatorname{Re} ({z^2}) = 0$ when it is given that $\operatorname{Re} (z) = \operatorname{Im} (z)$

Note:
As we know that $\omega ,{\omega ^2}$ also represent the complex number which are given by $\omega = \dfrac{{ - 1 + \sqrt 3 i}}{2}$
And ${\omega ^2} = \dfrac{{ - 1 - \sqrt 3 i}}{2}$ and we know that ${\omega ^2} = 1$ and hence we can say that ${\omega ^2} + \omega + 1 = 0$