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Question: Prove that if the complex numbers \[{z_1}\] and \[{z_2}\] with non-zero imaginary parts are such tha...

Prove that if the complex numbers z1{z_1} and z2{z_2} with non-zero imaginary parts are such that the product z1.z2{z_1}.{z_2} and the sum z1+z2{z_1} + {z_2} are real numbers, then z1{z_1} and z2{z_2} are conjugate complex numbers.

Explanation

Solution

Here we write z1{z_1} and z2{z_2} in expanded form with the help of general form of complex number i.e.z=x+iyz = 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.

  • i2=−1{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+iyz = x + iy is denoted by z‾\overline z and is given by taking the negative value of the imaginary part i.e. z‾=x−iy\overline z = x - iy.

Complete step-by-step answer:
Let us assume two complex numbers z1=x1+iy1{z_1} = {x_1} + i{y_1} and z2=x2+iy2{z_2} = {x_2} + i{y_2}.
Here y1,y2≠0{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 z1=x1+iy1{z_1} = {x_1} + i{y_1} and z2=x2+iy2{z_2} = {x_2} + i{y_2}
⇒z1.z2=(x1+iy1).(x2+iy2)\Rightarrow {z_1}.{z_2} = ({x_1} + i{y_1}).({x_2} + i{y_2})
Open the brackets on RHS of the equation.
⇒z1.z2=x1x2+ix1y2+ix2y1+i2y1y2\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 i2=−1{i^2} = - 1in RHS
⇒z1.z2=x1x2+ix1y2+ix2y1−y1y2\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
⇒z1.z2=(x1x2−y1y2)+i(x1y2+x2y1)\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 (x1x2−y1y2)({x_1}{x_2} - {y_1}{y_2})and the imaginary part is(x1y2+x2y1)({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.
⇒x1y2+x2y1=0\Rightarrow {x_1}{y_2} + {x_2}{y_1} = 0 … (1)
We calculate the sum of two complex numbers z1=x1+iy1{z_1} = {x_1} + i{y_1}andz2=x2+iy2{z_2} = {x_2} + i{y_2}
⇒z1+z2=(x1+iy1)+(x2+iy2)\Rightarrow {z_1} + {z_2} = ({x_1} + i{y_1}) + ({x_2} + i{y_2})
Open the brackets on RHS of the equation.
⇒z1+z2=x1+iy1+x2+iy2\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
⇒z1+z2=(x1+x2)+i(y1+y2)\Rightarrow {z_1} + {z_2} = ({x_1} + {x_2}) + i({y_1} + {y_2})
Here the real part is (x1+x2)({x_1} + {x_2})and the imaginary part is(y1+y2)({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.
⇒y1+y2=0\Rightarrow {y_1} + {y_2} = 0
Shift one of the values to the RHS of the equation.
⇒y1=−y2\Rightarrow {y_1} = - {y_2} … (2)
Substitute the value from equation (2) in equation (1)
⇒x1y2+x2(−y2)=0\Rightarrow {x_1}{y_2} + {x_2}( - {y_2}) = 0
⇒x1y2−x2y2=0\Rightarrow {x_1}{y_2} - {x_2}{y_2} = 0
Take y2{y_2}common on LHS of the equation
⇒y2(x1−x2)=0\Rightarrow {y_2}({x_1} - {x_2}) = 0
Since RHS is zero, this means either y2=0{y_2} = 0orx1−x2=0{x_1} - {x_2} = 0.
Since, we are given the question that both the complex numbers have non-zero imaginary parts.
We knowy2{y_2}is the imaginary part of the complex numberz2{z_2}.
Therefore,y2≠0{y_2} \ne 0
So we can write x1−x2=0{x_1} - {x_2} = 0
Shift the value of x2{x_2}to RHS of the equation.
⇒x1=x2\Rightarrow {x_1} = {x_2} … (3)
We have z1=x1+iy1{z_1} = {x_1} + i{y_1}
Use equation (2) to writey1=−y2{y_1} = - {y_2}and equation (3) to writex1=x2{x_1} = {x_2}in RHS ofz1=x1+iy1{z_1} = {x_1} + i{y_1}.
⇒z1=x2+i(−y2)\Rightarrow {z_1} = {x_2} + i( - {y_2})
⇒z1=x2−iy2\Rightarrow {z_1} = {x_2} - i{y_2}
Since, we knowz2=x2+iy2{z_2} = {x_2} + i{y_2}, then using the formula for conjugate of a complex number
z2‾=x2+iy2‾\overline {{z_2}} = \overline {{x_2} + i{y_2}}
We take the negative value of the imaginary term
z2‾=x2−iy2\overline {{z_2}} = {x_2} - i{y_2}
⇒z1=z2‾\Rightarrow {z_1} = \overline {{z_2}}
Therefore z1{z_1}and z2{z_2}are conjugate complex numbers.
Hence proved

Note: Students are likely to make mistake of not substituting the value of i2=−1{i^2} = - 1 in the solution and they try to group together the terms for imaginary part and they include the value with i2{i^2} in it, which is wrong. Keep in mind that value of i2=−1{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.