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Question: let \[{z_1}\] and \[{z_2}\] be two complex numbers such that \[{z_1} + {z_2}\] and \[{z_1}{z_2}\] bo...

let z1{z_1} and z2{z_2} be two complex numbers such that z1+z2{z_1} + {z_2} and z1z2{z_1}{z_2} both are real, then
(1) z1=−z2{z_1} = - {z_2}
(2) z1=bar z2{z_1} = {\text{bar }}{z_2}
(3) z1=−bar z2{z_1} = - {\text{bar }}{z_2}
(4) z1=z2{z_1} = {z_2}

Explanation

Solution

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 z1=a+ιb{z_1} = a + \iota b and z2=c+ιd{z_2} = c + \iota d be any two complex numbers. After that we will find z1+z2{z_1} + {z_2} and z1z2{z_1}{z_2} And then in the question it is given that z1+z2{z_1} + {z_2} and z1z2{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 z1=a+ιb{z_1} = a + \iota b and z2=c+ιd{z_2} = c + \iota d
Now first we will find z1+z2{z_1} + {z_2}
So, z1+z2=a+ιb+c+ιd{z_1} + {z_2} = a + \iota b + c + \iota d
⇒z1+z2=(a+c)+ι(b+d)\Rightarrow {z_1} + {z_2} = \left( {a + c} \right) + \iota \left( {b + d} \right)
Now, it is given that z1+z2{z_1} + {z_2} is real
It means its imaginary part is zero.
i.e., ι(b+d)=0\iota \left( {b + d} \right) = 0
either ι=0\iota = 0 or b+d=0b + d = 0 but ι\iota cannot be 00 because the value of ι\iota is −1\sqrt { - 1}
∴b+d=0\therefore b + d = 0
⇒d=−b −−−(1)\Rightarrow d = - b{\text{ }} - - - \left( 1 \right)
Now, we will find z1z2{z_1}{z_2}
So, z1z2=(a+ιb)(c+ιd){z_1}{z_2} = \left( {a + \iota b} \right)\left( {c + \iota d} \right)
On multiplying it, we get
z1z2=ac+ιad+ιbc+ι2bd{z_1}{z_2} = ac + \iota ad + \iota bc + {\iota ^2}bd
We know that ι2=−1{\iota ^2} = - 1
∴z1z2=ac+ιad+ιbc−bd\therefore {z_1}{z_2} = ac + \iota ad + \iota bc - bd
⇒z1z2=(ac−bd)+ι(ad+bc)\Rightarrow {z_1}{z_2} = \left( {ac - bd} \right) + \iota \left( {ad + bc} \right)
Now it is given that z1z2{z_1}{z_2} is real
which means its imaginary part is zero.
i.e., ι(ad+bc)=0\iota \left( {ad + bc} \right) = 0
either ι=0\iota = 0 or ad+bc=0ad + bc = 0 but ι\iota cannot be 00 because the value of ι\iota is −1\sqrt { - 1}
∴(ad+bc)=0\therefore \left( {ad + bc} \right) = 0
⇒ad=−bc\Rightarrow ad = - bc
But from (1)\left( 1 \right) , d=−bd = - b
∴\therefore above expression becomes,
a(−b)=−bca\left( { - b} \right) = - bc
On cancelling −b - b from both sides, we get
a=c −−−(2)a = c{\text{ }} - - - \left( 2 \right)
Now, z1=a+ιb{z_1} = a + \iota b
Using (1)\left( 1 \right) and (2)\left( 2 \right) , z1{z_1} can also be written as,
z1=c−ιd −−−(3){z_1} = c - \iota d{\text{ }} - - - \left( 3 \right)
Now if we see, z2=c+ιd{z_2} = c + \iota d
So, bar z2=c−ιd{\text{bar }}{z_2} = c - \iota d
∴\therefore equation (3)\left( 3 \right) can also be written as,
z1=bar z2{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 z1+z2=real{z_1} + {z_2} = {\text{real}} and z1z2=real{z_1}{z_2} = {\text{real}}
Now, by option (1) i.e., z1=−z2{z_1} = - {z_2}
So first we let z1=a+ιb{z_1} = a + \iota b
∴z2=−(a+ιb)\therefore {z_2} = - \left( {a + \iota b} \right)
So, z1+z2=a+ιb−(a+ιb){z_1} + {z_2} = a + \iota b - \left( {a + \iota b} \right)
⇒z1+z2=0=real\Rightarrow {z_1} + {z_2} = 0 = {\text{real}}
And z1z2=(a+ιb)(−a−ιb){z_1}{z_2} = \left( {a + \iota b} \right)\left( { - a - \iota b} \right)
⇒z1z2=−a2−2ιab+b2\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., z1=bar z2{z_1} = {\text{bar }}{z_2}
So first we let z1=a+ιb{z_1} = a + \iota b
∴z2=a−ιb\therefore {z_2} = a - \iota b
So, z1+z2=a+ιb+(a−ιb){z_1} + {z_2} = a + \iota b + \left( {a - \iota b} \right)
⇒z1+z2=2a=real\Rightarrow {z_1} + {z_2} = 2a = {\text{real}}
And z1z2=(a+ιb)(a−ιb){z_1}{z_2} = \left( {a + \iota b} \right)\left( {a - \iota b} \right)
⇒z1z2=a2+b2=real\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.