Question

Prove that if the complex numbers ${z_1}$ and ${z_2}$ with nonzero imaginary parts are such that the product ${z_1}, {z_2}$ and the sum ${z_1}$ + ${z_2}$ are real numbers, then ${z_1}$ and ${z_2}$ are conjugate complex numbers.

Answer 1

Hint: To solve this question, we have to remember some basic concepts of complex numbers. We will use the concept that for a complex number,
A number in the form of $a + ib$ where $a$ and $b$ are real numbers, is called a complex number.
For the complex number $z = a + ib$, $a$ is the real part and denoted by Re$z$ and $b$ is the imaginary part denoted by Im$z$ of the complex number.

Complete step-by-step answer:
According to the question,
We know that, if a complex number is equal to $0$, then its real part and imaginary part both will be equal to zero.
Let $z = a + ib$, where $a$ and $b$ both are equal to$0$,
Let${z_1}$ and ${z_2}$ be two complex numbers
${z_1} = x + iy$
${z_2} = a + ib$ Where $\operatorname{Im} z = b,\operatorname{Re} z = a$
We are finding the product of both complex numbers.
${z_1}{z_2} = (x + iy)(a + ib)$
Open the brackets by multiplying terms
${z_1}{z_2} = xa + ixb + iay + {i^2}b$
${z_1}{z_2}$$ = (xa - b) + i(xb + ay)............(1)$
We need to add the both complex numbers
∴${z_1} + {z_2} = (x + a) + i(y + b)........(2)$
${z_1}{z_2}\& {z_1} + {z_2}$ are real numbers.
First put the imaginary part of eqn. $(1)$ equals to zero
$xb + ay = 0...\left( 3 \right)$
Secondly put the imaginary part of eqn. (2) equals to zero
$y + b = 0$
$y = - b.......(4)$
Put $y = - b$ in equation $\left( 3 \right)$ we get,
$xb - ab = 0$
$b(x - a) = 0$
$\therefore x = a..........(5)$
Putting the values of $a$ and $b$ in the above complex number, ${z_1}$ and ${z_2}$ we get,
${z_1} = a - ib$
${z_2} = a + ib$
Here ${z_1}{\text{ and }}{z_2}$ are complex conjugates.

Note: The chances of mistakes are if the conjugate of number is not taken correctly or there may be mistakes during the interpretation of the conditions given in the question.
One should not get confused with modulus and the number.