Question

Statement 1: If $z_1+z_2=a$ and $z_1{z}_2=b$, where $a=a$ and $b=\overline{b}$, then $arg\left(z_1{z}_2\right)=0$.
Statement 2: The sum and product of two complex numbers are real if and only if they are conjugate of each other.

• A. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.
• B. Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.
• C. Statement 1 is true and Statement 2 is false.
• D. Statement 1 is false and Statement 2 is true.

Answer 1

The correct option is A Both the statements are true, and Statement 2 is the correct explanation for Statement 1.

Let the two complex numbers be conjugate of each other. Let complex numbers be $z_1=x+iy$ and $z_2=x-iy$.

Then, $z_1+z_2=(x+iy)+(x-ix)=2x$, which is real and $z_1{z}_2=(x+iy)(x-iy)=x^2-i^2{y}^2=x^2+y^2$, which is real.

Conversely, let $z_1$ and $z_2$ be two complex numbers such that their sum $z_1+z_2$ and product $z_1{z}_2$ both are real.

Let $z_1=x_1+i{y}_1$ and $z_2=x_2+i{y}_2$.

Then,
$z_1+z_2=(x_1+x_2)+i(y_1+y_2)$

and $z_1{z}_2=(x_{1}x{s}_2-y_1{y}_2)+i(x_1{y}_2+x_2{y}_1)$

Now, $z_1+z_2$ and $z_1{z}_3$ are real. Hence,

$b_1+b_2=0$ and $a_1{b}_2+a_2{b}_1=0[\therefore z$ is real $\Rightarrow Im\left(z\right)=0]$

$\Rightarrow b_2=-b_1$ and n$a_1{b}_2+a_2{b}_1=0$

$=-b_1$ and $-a_1{b}_1+a_2{b}_1=0$

$=-b_1$ and $(a_2-a_1)b_1=0$

$=-b_1$ and $a_2-a_1=0$

$=-b_1$ and $a_2=a_1$

$\Rightarrow z_2=a_2+i{b}_2=a_1-i{b}_1$

$={\overline{z}}_1$

Hence, $z_1$ and $z_2$ are conjugate of each other. Hence, statement 2 is true,

Also in statement 1, $a=\overline{a}$ and $b=\overline{b}$, then a and b are real.

Thus, $z_1+z_2$ and $z_1{z}_2$ are real. So,

$z_2={\overline{z}}_2$

$\Rightarrow arg\left(z_1{z}_2\right)=arg\left(z_1{\overline{z}}_1\right)=arg\left(|z_1{|}^2\right)=0$

Hence, Statement 1 is correct and statement 2 is correct explanation of statement 1.