Question

If $\left|z_{1}\right|=\left|z_{2}\right|$ and $\arg \left(z_{1}\right)+\arg \left(z_{2}\right)=\pi / 2$, then

• A. $z_{1} z_{2}$ is purely real
• B. $z_{1} z_{2}$ is purely imaginary
• C. $\left(z_{1}+z_{2}\right)^{2}$ is purely real.
• D. $\arg \left(z_{1}^{-1}\right)+\arg \left(z_{2}^{-1}\right)=\frac{\pi}{2}$

Answer 1

Answer: (B)

Let $\left|z_{1}\right|=\left|z_{2}\right|=r$
$\Rightarrow z_{1}=r(\cos \theta+i \sin \theta) $
and $z_{2}=r\left(\cos \left(\frac{\pi}{2}-\theta\right)+i \sin \left(\frac{\pi}{2}-\theta\right)\right)$
$\Rightarrow z_{1} z_{2}=r^{2} i$, which is purely imaginary
$ z_{1}+z_{2}=r[(\cos \theta+\sin \theta)+i(\cos \theta+\sin \theta)] $
$\Rightarrow \left(z_{1}+z_{2}\right)^{2}=2 r^{2} \cdot(\cos \theta+\sin \theta)^{2} \cdot i$
which is purely imaginary.
Also $\arg \left(z_{1}^{-1}\right)+\arg \left(z_{2}^{-1}\right)=-\frac{\pi}{2}$.