Question

Let $z_{1}$ and $z_{2}$ are two complex numbers such that $\left|z_{1}\right|=\left|z_{2}\right|$ and $arg\left(z_{1}\right)+arg\left(z_{2}\right)=\pi $ , then $z_{1}$ equals to

• A. $z_{2}$
• B. $-z_{2}$
• C. $\overline{z}_{2}$
• D. $-\overline{z}_{2}$

Answer 1

Answer: (D)

We have given that,
$\left|z_{1}\right|=\left|z_{2}\right|$ and $arg\left(z_{1}\right)=\pi -arg\left(z_{2}\right)$
$z_{2}=re^{i \theta },\bar{z}_{2}=re^{- i \theta }$
$z_{1}=re^{i \left(\right. \pi - \theta \left.\right)}=re^{i \pi }\cdot e^{- i \theta }$
$=r\left(\right.cos\pi +isin\pi \left.\right)\cdot e^{- i \theta }$
$=-re^{- i \theta }=-\overline{z}_{2}$
Alternate Solution

From figure, it is clear that $z_{1}=-\bar{z}_{2}$