Question

If $z_1,z_2$ are two complex numbers such that $arg(z_1+z_2)=0$ and $Im\left(z_1{z}_2\right)=0$, then.

• A. $z_1=-z_2$
• B. $z_1=z_2$
• C. $z_1=\overrightarrow{z_2}$
• D. none of these

Answer 1

The correct option is C $z_1=\overrightarrow{z_2}$

Let $z_1=a+ib$

and $z_2=c+id$

Now $z_1+z_2=(a+c)+i(b+d)$

Given $arg(z_1+z_2)=0$

$\Rightarrow ta{n}^{-1}\left[\frac{b+d}{a+c}\right]=0$

$\Rightarrow b+d=0$.........$\left(1\right)$

Now,

$z_1{z}_2=(ac-bd)+i(ad+bc)$

Given $Im\left(z_1{z}_2\right)=0$

$\Rightarrow ad+bc=0$.........$\left(2\right)$

Putting $b=-d$ from $\left(1\right)$ in $\left(2\right)$ we get $a=c$

Hence $z_1=a+ib=c-id=$conjugate of $z_2=z_2^{\to }$

Hence option C is correct.