Question

Prove that if the complex numbers \[{z_1}\] and \[{z_2}\] with non-zero 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: Here we write \[{z_1}\] and \[{z_2}\] in expanded form with the help of general form of complex number i.e.\[z = x + iy\]. Calculate the product and sum of these two complex numbers by using the expanded forms. Equate the imaginary parts obtained in both sum and product to zero and write the relation between real parts of two complex numbers and imaginary parts of two complex numbers.
* \[{i^2} = - 1\]
* A complex number is said to be purely real if it has an imaginary part equal to zero and purely imaginary if it has a real part equal to zero.
* Conjugate of a complex number \[z = x + iy\] is denoted by \[\overline z \] and is given by taking the negative value of the imaginary part i.e. \[\overline z = x - iy\].

Complete step-by-step answer:
Let us assume two complex numbers \[{z_1} = {x_1} + i{y_1}\] and \[{z_2} = {x_2} + i{y_2}\].
Here \[{y_1},{y_2} \ne 0\] as the imaginary part of the complex numbers is given as non-zero.
We calculate the product of two complex numbers \[{z_1} = {x_1} + i{y_1}\] and \[{z_2} = {x_2} + i{y_2}\]
\[ \Rightarrow {z_1}.{z_2} = ({x_1} + i{y_1}).({x_2} + i{y_2})\]
Open the brackets on RHS of the equation.
\[ \Rightarrow {z_1}.{z_2} = {x_1}{x_2} + i{x_1}{y_2} + i{x_2}{y_1} + {i^2}{y_1}{y_2}\]
Substitute the value of \[{i^2} = - 1\]in RHS
\[ \Rightarrow {z_1}.{z_2} = {x_1}{x_2} + i{x_1}{y_2} + i{x_2}{y_1} - {y_1}{y_2}\]
Group the real part and the imaginary part on RHS
\[ \Rightarrow {z_1}.{z_2} = ({x_1}{x_2} - {y_1}{y_2}) + i({x_1}{y_2} + {x_2}{y_1})\]
Here the real part is \[({x_1}{x_2} - {y_1}{y_2})\]and the imaginary part is\[({x_1}{y_2} + {x_2}{y_1})\].
Since, we know the product of two given complex numbers is a real number, so the imaginary part must be equal to zero.
\[ \Rightarrow {x_1}{y_2} + {x_2}{y_1} = 0\] … (1)
We calculate the sum of two complex numbers \[{z_1} = {x_1} + i{y_1}\]and\[{z_2} = {x_2} + i{y_2}\]
\[ \Rightarrow {z_1} + {z_2} = ({x_1} + i{y_1}) + ({x_2} + i{y_2})\]
Open the brackets on RHS of the equation.
\[ \Rightarrow {z_1} + {z_2} = {x_1} + i{y_1} + {x_2} + i{y_2}\]
Group the real part and the imaginary part on RHS
\[ \Rightarrow {z_1} + {z_2} = ({x_1} + {x_2}) + i({y_1} + {y_2})\]
Here the real part is \[({x_1} + {x_2})\]and the imaginary part is\[({y_1} + {y_2})\].
Since, we know the product of two given complex numbers is a real number, so the imaginary part must be equal to zero.
\[ \Rightarrow {y_1} + {y_2} = 0\]
Shift one of the values to the RHS of the equation.
\[ \Rightarrow {y_1} = - {y_2}\] … (2)
Substitute the value from equation (2) in equation (1)
\[ \Rightarrow {x_1}{y_2} + {x_2}( - {y_2}) = 0\]
\[ \Rightarrow {x_1}{y_2} - {x_2}{y_2} = 0\]
Take \[{y_2}\]common on LHS of the equation
\[ \Rightarrow {y_2}({x_1} - {x_2}) = 0\]
Since RHS is zero, this means either \[{y_2} = 0\]or\[{x_1} - {x_2} = 0\].
Since, we are given the question that both the complex numbers have non-zero imaginary parts.
We know\[{y_2}\]is the imaginary part of the complex number\[{z_2}\].
Therefore,\[{y_2} \ne 0\]
So we can write \[{x_1} - {x_2} = 0\]
Shift the value of \[{x_2}\]to RHS of the equation.
\[ \Rightarrow {x_1} = {x_2}\] … (3)
We have \[{z_1} = {x_1} + i{y_1}\]
Use equation (2) to write\[{y_1} = - {y_2}\]and equation (3) to write\[{x_1} = {x_2}\]in RHS of\[{z_1} = {x_1} + i{y_1}\].
\[ \Rightarrow {z_1} = {x_2} + i( - {y_2})\]
\[ \Rightarrow {z_1} = {x_2} - i{y_2}\]
Since, we know\[{z_2} = {x_2} + i{y_2}\], then using the formula for conjugate of a complex number
\[\overline {{z_2}} = \overline {{x_2} + i{y_2}} \]
We take the negative value of the imaginary term
\[\overline {{z_2}} = {x_2} - i{y_2}\]
\[ \Rightarrow {z_1} = \overline {{z_2}} \]
Therefore \[{z_1}\]and \[{z_2}\]are conjugate complex numbers.
Hence proved

Note: Students are likely to make mistake of not substituting the value of \[{i^2} = - 1\] in the solution and they try to group together the terms for imaginary part and they include the value with \[{i^2}\] in it, which is wrong. Keep in mind that value of \[{i^2} = - 1\] so it is completely real. Also, keep a check on the change of sign when shifting values from one side of the equation to another side of the equation.