Question

let \[{z_1}\] and \[{z_2}\] be two complex numbers such that \[{z_1} + {z_2}\] and \[{z_1}{z_2}\] both are real, then
(1) \[{z_1} = - {z_2}\]
(2) \[{z_1} = {\text{bar }}{z_2}\]
(3) \[{z_1} = - {\text{bar }}{z_2}\]
(4) \[{z_1} = {z_2}\]

Answer 1

Hint: To solve this question, we will use the concept of the real and imaginary part of complex numbers i.e., if it is given that a complex number is real, it means that its imaginary part is zero. So, first we let \[{z_1} = a + \iota b\] and \[{z_2} = c + \iota d\] be any two complex numbers. After that we will find \[{z_1} + {z_2}\] and \[{z_1}{z_2}\] And then in the question it is given that \[{z_1} + {z_2}\] and \[{z_1}{z_2}\] both are real so using the above mentioned concept we will find the required result by doing some simplifications.

Complete step by step answer:
Let \[{z_1} = a + \iota b\] and \[{z_2} = c + \iota d\]
Now first we will find \[{z_1} + {z_2}\]
So, \[{z_1} + {z_2} = a + \iota b + c + \iota d\]
\[ \Rightarrow {z_1} + {z_2} = \left( {a + c} \right) + \iota \left( {b + d} \right)\]
Now, it is given that \[{z_1} + {z_2}\] is real
It means its imaginary part is zero.
i.e., \[\iota \left( {b + d} \right) = 0\]
either \[\iota = 0\] or \[b + d = 0\] but \[\iota \] cannot be \[0\] because the value of \[\iota \] is \[\sqrt { - 1} \]
\[\therefore b + d = 0\]
\[ \Rightarrow d = - b{\text{ }} - - - \left( 1 \right)\]
Now, we will find \[{z_1}{z_2}\]
So, \[{z_1}{z_2} = \left( {a + \iota b} \right)\left( {c + \iota d} \right)\]
On multiplying it, we get
\[{z_1}{z_2} = ac + \iota ad + \iota bc + {\iota ^2}bd\]
We know that \[{\iota ^2} = - 1\]
\[\therefore {z_1}{z_2} = ac + \iota ad + \iota bc - bd\]
\[ \Rightarrow {z_1}{z_2} = \left( {ac - bd} \right) + \iota \left( {ad + bc} \right)\]
Now it is given that \[{z_1}{z_2}\] is real
which means its imaginary part is zero.
i.e., \[\iota \left( {ad + bc} \right) = 0\]
either \[\iota = 0\] or \[ad + bc = 0\] but \[\iota \] cannot be \[0\] because the value of \[\iota \] is \[\sqrt { - 1} \]
\[\therefore \left( {ad + bc} \right) = 0\]
\[ \Rightarrow ad = - bc\]
But from \[\left( 1 \right)\] , \[d = - b\]
\[\therefore \] above expression becomes,
\[a\left( { - b} \right) = - bc\]
On cancelling \[ - b\] from both sides, we get
\[a = c{\text{ }} - - - \left( 2 \right)\]
Now, \[{z_1} = a + \iota b\]
Using \[\left( 1 \right)\] and \[\left( 2 \right)\] , \[{z_1}\] can also be written as,
 \[{z_1} = c - \iota d{\text{ }} - - - \left( 3 \right)\]
Now if we see, \[{z_2} = c + \iota d\]
So, \[{\text{bar }}{z_2} = c - \iota d\]
\[\therefore \] equation \[\left( 3 \right)\] can also be written as,
\[{z_1} = {\text{bar }}{z_2}\]
Hence, option (2) is the correct answer.

Note: While solving this question, we can also find the solution by using the given options.
Like, it is given that \[{z_1} + {z_2} = {\text{real}}\] and \[{z_1}{z_2} = {\text{real}}\]
Now, by option (1) i.e., \[{z_1} = - {z_2}\]
So first we let \[{z_1} = a + \iota b\]
\[\therefore {z_2} = - \left( {a + \iota b} \right)\]
So, \[{z_1} + {z_2} = a + \iota b - \left( {a + \iota b} \right)\]
\[ \Rightarrow {z_1} + {z_2} = 0 = {\text{real}}\]
And \[{z_1}{z_2} = \left( {a + \iota b} \right)\left( { - a - \iota b} \right)\]
\[ \Rightarrow {z_1}{z_2} = - {a^2} - 2\iota ab + {b^2}\] which is not real.
Thus, both the conditions are not satisfied.
Hence, option (1) is not correct.
Now, by option (2) i.e., \[{z_1} = {\text{bar }}{z_2}\]
So first we let \[{z_1} = a + \iota b\]
\[\therefore {z_2} = a - \iota b\]
So, \[{z_1} + {z_2} = a + \iota b + \left( {a - \iota b} \right)\]
\[ \Rightarrow {z_1} + {z_2} = 2a = {\text{real}}\]
And \[{z_1}{z_2} = \left( {a + \iota b} \right)\left( {a - \iota b} \right)\]
\[ \Rightarrow {z_1}{z_2} = {a^2} + {b^2} = {\text{real}}\]
Thus, both the conditions are satisfied.
Hence, option (2) is correct which is the required answer.
But it will be way too long, so try not to use it.