$\frac{1 + 7i}{(2 - i)^2} = $

  • A
    $\sqrt{2} \left( \cos \frac{3\pi}{4} + i\sin \frac{3\pi}{4} \right)$
  • B
    $\sqrt{2} \left( \cos \frac{\pi}{4} + i\sin \frac{\pi}{4} \right)$
  • C
    $\left( \cos \frac{3\pi}{4} + i\sin \frac{3\pi}{4} \right)$
  • D
    None of these

Explore More

Similar Questions

Let $S = \{z \in \mathbb{C} : z^2 + 4z + 16 = 0\}$. Then $\sum_{z \in S} |z + \sqrt{3}i|^2$ is equal to:

The number of solutions of the equation $z^2 + \bar{z} = 0$ is

Let $S$ be the set of all $(\alpha, \beta)$ such that $\pi < \alpha, \beta < 2\pi$,for which the complex number $\frac{1-i \sin \alpha}{1+2i \sin \alpha}$ is purely imaginary and $\frac{1+i \cos \beta}{1-2i \cos \beta}$ is purely real. Let $Z_{\alpha \beta} = \sin 2\alpha + i \cos 2\beta$ for $(\alpha, \beta) \in S$. Then $\sum_{(\alpha, \beta) \in S} \left(i Z_{\alpha \beta} + \frac{1}{i \bar{Z}_{\alpha \beta}}\right)$ is equal to:

If $f(x)$ is a polynomial of degree $n$ with rational coefficients and $1+2i, 2-\sqrt{3}$ and $5$ are three roots of $f(x)=0$,then the least value of $n$ is

The set of all values of $\theta$ such that $\frac{1-i \cos \theta}{1+2 i \sin \theta}$ is purely imaginary is

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