The amplitude (argument) of $(1+i)^{5}$ is

  • A
    $\frac{3 \pi}{4}$
  • B
    $\frac{-3 \pi}{4}$
  • C
    $\frac{-5 \pi}{4}$
  • D
    $\frac{5 \pi}{4}$

Explore More

Similar Questions

If $z = \frac{(2-i)(1+i)^3}{(1-i)^2}$,then $\operatorname{Arg}(z) = $

Let $z_{1}$ and $z_{2}$ be two complex numbers such that $\overline{z}_{1} = i \overline{z}_{2}$ and $\arg \left( \frac{z_{1}}{\overline{z}_{2}} \right) = \pi$. Then:

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:

If $arg(z) < 0$,then $arg(-z) - arg(z)$ equals

If $z$ is a complex number such that $|z| = 4$ and $\text{arg}(z) = \frac{5\pi}{6}$,then $z$ is equal to

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo