If z = left(frac{sqrt{3}+i}{2}right)^5 + left | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $\omega = \frac{\sqrt{3}+i}{2} = \cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right) = e^{i\pi/6}$.Then the given expression is $

Vedclass · Accepted Answer

$\operatorname{Im}(z) = 0$

Vedclass · Answer

(D) Let $\omega = \frac{\sqrt{3}+i}{2} = \cos\left(\frac{\pi}{6}ight) + i\sin\left(\frac{\pi}{6}ight) = e^{i\pi/6}$.Then the given expression is $z = (e^{i\pi/6})^5 + (e^{-i\pi/6})^5$.$z = e^{i5\pi/6} + e^{-i5\pi/6}$.Using the identity $e^{i	heta} + e^{-i	heta} = 2\cos(	heta)$, we get $z = 2\cos\left(\frac{5\pi}{6}ight)$.Since $\cos\left(\frac{5\pi}{6}ight) = -\frac{\sqrt{3}}{2}$, we have $z = 2\left(-\frac{\sqrt{3}}{2}ight) = -\sqrt{3}$.Since $z = -\sqrt{3} + 0i$, the imaginary part $\operatorname{Im}(z) = 0$.

If $z = \left(\frac{\sqrt{3}+i}{2}\right)^5 + \left(\frac{\sqrt{3}-i}{2}\right)^5$, then

Similar Questions

$\left[\frac{1+\cos \left(\frac{\pi}{12}\right)+i \sin \left(\frac{\pi}{12}\right)}{1+\cos \left(\frac{\pi}{12}\right)-i \sin \left(\frac{\pi}{12}\right)}\right]^{72}=$

For any real number $n \in \mathbb{R}$,$(\cosh x + \sinh x)^n =$

If $(\sqrt{3}+i)^{100}=2^{99}(p+iq)$,then $p$ and $q$ are roots of the equation :

If $\cos \theta + i \sin \theta, \theta \in R$, is a root of the equation $a_0 x^n + a_1 x^{n-1} + \ldots + a_{n-1} x + a_n = 0$, where $a_0, a_1, \ldots, a_n \in R$ and $a_0 \neq 0$, then the value of $a_1 \sin \theta + a_2 \sin 2 \theta + \ldots + a_n \sin n \theta$ is:

$\sum_{r=1}^{16}\left(\sin \frac{2 r \pi}{17}+i \cos \frac{2 r \pi}{17}\right)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module