Question

Let $z_1,z_2$ two complex numbers such that $z_1+z_2$ and $z_1{z}_2$ both are real, then

• A. $z_1=-z_2$
• B. $z_1=\overline{z_2}$
• C. $z_1=-\overline{z_2}$
• D. $z_1=z_2$

Answer 1

Answer: (B)

$z_1=\overline{z_2}$

Given

$z_1+z_2$=real

$z_1{z}_2$ = real

Solve by options

$\left(A\right)=z_1=-z_2$

assume $z_1=n+iy$

$z_2=-(n+iy)$

$z_1+z_2=n+iy[-(n+iy\left)\right]$

$=0$ ( real )

$z_1.z_2$ = (n+y)9-(n+iy)$

$=[n^2-y^2+2ixy]$

Not real

So, option A is not correct

$\left(B\right)2_1=z_2$

assume $z_1=n+iy$

$z_2=n-iy$

Now $z_1+z_2=n+iy+n-iy$

$2n$ ( real )

$z_1{z}_2$ = $(n+iy)(n-iy)$

$n^2+y^2$ (real )

option B satifies all condation so option B is correct and Answer is B.