Question

Let $z_1,z_2$ be 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

The correct option is B

$z_1=\overline{z_2}$

Explanation for the correct answer:

Calculating $z_1$:

Given, $z_1+z_{2}and{z}_1{z}_2$ are real.

Let, $z_1=a+ib,z_2=c+id$

Since, $z_1+z_2$ is real,

$$\begin{gathered} \Rightarrow (a+c)+i(b+d)isreal \\ \Rightarrow b+d=0\left[z_1+z_{2}isreal\right] \\ \Rightarrow b=-d......\left(1\right) \end{gathered}$$

Also, $z_1{z}_2$ is real ,

$$\begin{gathered} =(a+ib).(c+id)isreal \\ =ac+iad+ibc+i^{2}bdisreal \\ =ac+i\left(ad+bc\right)-bdisreal\mathbf{[}\mathbf{\because }{\mathit{i}}^{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{1}] \\ =(ac-bd)+i(ad+bc)isreal \\ \Rightarrow ad+bc=0\mathbf{[}{\mathbf{z}}_{\mathbf{1}\mathbf{.}}{\mathbf{z}}_{\mathbf{2}}\mathbf{}\mathbf{i}\mathbf{s}\mathbf{}\mathbf{r}\mathbf{e}\mathbf{a}\mathbf{l}\mathbf{]} \\ \Rightarrow a(-b)+bc=0\mathbf{[}\mathit{F}\mathit{r}\mathit{o}\mathit{m}\mathbf{}\mathbf{(}\mathbf{1}\mathbf{\left)}\right] \\ \Rightarrow a=c.....\left(2\right) \end{gathered}$$

Therefore,

$$\begin{gathered} z_1=a+ib \\ =c-id\mathbf{[}\mathit{F}\mathit{r}\mathit{o}\mathit{m}\mathbf{}\mathbf{(}\mathbf{1}\mathbf{)}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathbf{(}\mathbf{2}\mathbf{\left)}\mathbf{\right]} \\ =\overline{z_2}\mathbf{[}\mathbf{\because }{\mathit{z}}_{\mathbf{2}}\mathbf{=}\mathit{c}\mathbf{}\mathbf{+}\mathbf{}\mathit{i}\mathit{d}\mathbf{]}\mathbf{} \end{gathered}$$

Hence, the correct answer is option (B).