If $z$ and $w$ are two complex numbers such that $|zw| = 1$ and $\arg(z) - \arg(w) = \frac{\pi}{2}$,then
- A$\bar{z}w = i$
- B$z\bar{w} = \frac{-1 + i}{\sqrt{2}}$
- C$z\bar{w} = \frac{1 - i}{\sqrt{2}}$
- D$\bar{z}w = -i$
Explore More
Similar Questions
Let $z \in \mathbb{C}$ and $i=\sqrt{-1}$. If $a, b, c \in (0,1)$ are such that $a^2+b^2+c^2=1$ and $b+ic=(1+a)z$,then $\frac{1+iz}{1-iz}=$
If $\frac{2z_1}{3z_2}$ is a purely imaginary number,then $\left| \frac{z_1 - z_2}{z_1 + z_2} \right| =$
Medium
View SolutionThe biquadratic equation,two of whose roots are $1+i$ and $1-\sqrt{2}$,is
If ${x_n} = \cos \left( \frac{\pi }{3^n} \right) + i\sin \left( \frac{\pi }{3^n} \right)$,then the product ${x_1} \cdot {x_2} \cdot {x_3} \cdots {x_\infty }$ is equal to:
Difficult
View SolutionThe value of the infinite product $(\cos \theta + i\sin \theta )(\cos \frac{\theta }{2} + i\sin \frac{\theta }{2})(\cos \frac{\theta }{2^2} + i\sin \frac{\theta }{2^2}) \dots$ is
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo