Question

If Z and Z’ are complex numbers such that $Z.Z'=Z$ then $Z'=$ ?
(a) $0+i0$
(b) $1+0i$
(c) $0+i$
(d) $1+i$

Answer 1

Hint: In this question, we will write Z the complex number in general form Z = a + bi, where i is the root of -1 or known as the ‘iota’. Z’ is known as the conjugate of the complex number and is found by making the signs opposite of a complex number. To solve this question we have to put Z = a + bi and Z’ = a – bi. Substitute in the expression $Z.Z'=Z$ and solve for Z.

Complete step by step answer:
In the given question, we have,
$Z.Z'=Z\ldots (i)$ , where Z is complex number and Z’ is conjugate of complex numbers.
Let us consider, $Z=a+ib$ , where a and b are real numbers.
Then Z’ is a conjugate number of Z.
Now, we know that the complex conjugate of a complex number is a complex number with equal real part and imaginary part equal in magnitude but opposite in sign.
Therefore, the real part of Z’ here will be the same as of Z, that is a. And, the imaginary part will be equal in magnitude but opposite in sign, that is -b.
Hence, $Z'=a-ib$
Now, dot product of Z and Z’ can be given as,
$Z.Z'=\left( a+ib \right)\left( a-ib \right)$
Applying distributive law, we get,
$\begin{align}
  & Z.Z'=a.a-a.ib+ib.a-ib.ib \\
 & ={{a}^{2}}-iab+iba-{{b}^{2}}{{i}^{2}} \\
\end{align}$
Here, $iab=iba$ and $i'=-1$ therefore,
$\begin{align}
  & Z.Z'={{a}^{2}}-iab+iab-{{b}^{2}}\left( -1 \right) \\
 & ={{a}^{2}}+{{b}^{2}} \\
\end{align}$
Now, using this value of Z.Z’ in equation (i), we get,
$\begin{align}
  & Z.Z'=Z \\
 & \Rightarrow {{a}^{2}}+{{b}^{2}}=a-ib \\
\end{align}$
Here, a and b are real numbers, So ${{a}^{2}}+{{b}^{2}}$ is also real and Z.Z’ have no imaginary part .
So, we can write,
$\left( {{a}^{2}}+{{b}^{2}} \right)+i.0=a+i.\left( -b \right)$
Comparing real part and imaginary part, we get,
${{a}^{2}}+{{b}^{2}}=a$ and $-b=0$
Substituting $b=0$ in first equation, we get,
$\begin{align}
  & {{a}^{2}}+{{0}^{2}}=a \\
 & \Rightarrow {{a}^{2}}=a \\
\end{align}$
Subtracting a from both sides of the equation, we get,
$\Rightarrow {{a}^{2}}-a=0$
Taking a common we get,
$a\left( a-1 \right)=0$
Therefore, $a=0$ or $a-1=0$
$\Rightarrow a=0$ or $a=1$
Hence, the required number will be,
$0+0i$ and $1+0i$
Therefore, the correct answer is option (a) and (b).

Note:
The possible mistake that can occur in this question is that we forget to take the opposite sign of the complex number in the conjugate of Z, this will give the final answer incorrect. So it is important to change the sign of the complex number in the conjugate.