Class 11Mathematics1 Complex Numbers and Quadratic Equations: Complex NumbersArgument of complex numbers
EasyIf the amplitude of $(z-1-2i)$ is $\frac{\pi}{3}$,then the locus of $z$ is
- A$y=\sqrt{3}x+(2-\sqrt{3})$
- B$y=\sqrt{3}x-\sqrt{3}$
- C$x=\sqrt{3}y+(2-\sqrt{3})$
- D$y=\sqrt{3}x+2$
Explore More
Similar Questions
Let the two values of $z = \sqrt{\frac{1-i}{1+i}}$ be $z_1$ and $z_2$. If $-\frac{\pi}{2} < \operatorname{Arg}(z_1) < \operatorname{Arg}(z_2) < \pi$,then $\arg(z_1) + \arg(z_2) = $
DifficultAP EAMCET 2023
View SolutionIf $\frac{\pi}{2} < \alpha < \frac{3\pi}{2}$,then the modulus and argument of $(1 + \cos 2\alpha) + i \sin 2\alpha$ are respectively:
Match the items of List-$I$ with those of List-$II$:
List-$I$ (Complex number) List-$II$ (Polar form) $(i) \sqrt{3}-i$ $(a) 2 \operatorname{cis} \frac{\pi}{6}$ $(ii) \sqrt{3}+i$ $(b) 2 \operatorname{cis} \frac{5 \pi}{6}$ $(iii) -\sqrt{3}+i$ $(c) 2 \operatorname{cis}\left(-\frac{5 \pi}{6}\right)$ $(iv) -\sqrt{3}-i$ $(d) 2 \operatorname{cis}\left(-\frac{\pi}{6}\right)$
The correct matching is:
| List-$I$ (Complex number) | List-$II$ (Polar form) |
| $(i) \sqrt{3}-i$ | $(a) 2 \operatorname{cis} \frac{\pi}{6}$ |
| $(ii) \sqrt{3}+i$ | $(b) 2 \operatorname{cis} \frac{5 \pi}{6}$ |
| $(iii) -\sqrt{3}+i$ | $(c) 2 \operatorname{cis}\left(-\frac{5 \pi}{6}\right)$ |
| $(iv) -\sqrt{3}-i$ | $(d) 2 \operatorname{cis}\left(-\frac{\pi}{6}\right)$ |
The correct matching is:
The complex number with argument $\frac{5 \pi}{6}$ at a distance of $2$ units from the origin is
If $z = \frac{-2}{1 + \sqrt{3}i}$,then the value of $arg(z)$ is
Easy
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