Question

If $z \ne 0$ is a complex number, then prove that $\operatorname{Re} \left( z \right) = 0 \Rightarrow \operatorname{Im} \left( {{z^2}} \right) = 0$.

Answer 1

Hint: First of all let the complex number be $z = x + iy$, where $x$ is the real part and $y$ is the imaginary part. Now, here we are given that the real part is 0. Therefore, put $x = 0$. Now, square the complex number and we will get the imaginary part of ${z^2}$ as 0.

Complete step by step solution:
In this question, we are given that $z$ is a complex number and it is not equal to 0. Also we are given that the real part of this complex number $z$ is equal to 0 and we need to prove that the imaginary part of the given complex number square is equal to 0.
So, first of all a complex number is a number that contains imaginary numbers and real numbers. The imaginary part is denoted by $i$.
$i = \sqrt { - 1} $
Let $z = x + iy$ be our complex number.
Now, here x is the real part of the complex number and y is the imaginary part.
Now, we are given that the real part of z that is $\operatorname{Re} \left( z \right)$ is equal to 0. Therefore,
 $
   \Rightarrow z = 0 + iy \\
   \Rightarrow z = iy \\
 $
Now, we need to prove that the imaginary part of ${z^2}$ is equal to zero. So, squaring the above equation, we get
$ \Rightarrow {z^2} = {i^2}{y^2}$ - - - - - - (1)
Now, we know that
$
   \Rightarrow i = \sqrt { - 1} \\
   \Rightarrow {i^2} = - 1 \\
 $
Therefore, equation (1) becomes
$ \Rightarrow {z^2} = - {y^2}$
Here, we can see that there is no imaginary part in the above equation.
Hence, we can say that $\operatorname{Im} \left( {{z^2}} \right) = 0$.

Note:
Note that when there is no imaginary part in a complex number, then that number will be purely real. For example: $z = x + y$ is a purely real number as there is no imaginary number.
When there is no real part in a given complex number, then the number is called a purely imaginary number. For example: $z = 6i$ is a purely imaginary number as there is no real number.