यदि z = frac{sqrt{3} + i}{2} है,तो left(z^{10 | vedclass

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

Question

(C) दिया है,$z = \frac{\sqrt{3} + i}{2} = \cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right)$.डी मॉइवर प्रमेय का उपयोग करते हुए,$z^{1

Vedclass · Accepted Answer

$i$

Vedclass · Answer

(C) दिया है,$z = \frac{\sqrt{3} + i}{2} = \cos \left(\frac{\pi}{6}ight) + i \sin \left(\frac{\pi}{6}ight)$.डी मॉइवर प्रमेय का उपयोग करते हुए,$z^{101} = \cos \left(\frac{101\pi}{6}ight) + i \sin \left(\frac{101\pi}{6}ight)$.चूंकि $\frac{101\pi}{6} = 17\pi - \frac{\pi}{6}$,इसलिए $z^{101} = \cos \left(17\pi - \frac{\pi}{6}ight) + i \sin \left(17\pi - \frac{\pi}{6}ight) = -\cos \left(\frac{\pi}{6}ight) + i \sin \left(\frac{\pi}{6}ight) = \frac{-\sqrt{3} + i}{2}$.साथ ही,$i^{103} = i^{100} 	imes i^3 = 1 	imes (-i) = -i$.अतः,$z^{101} + i^{103} = \frac{-\sqrt{3} + i}{2} - i = \frac{-\sqrt{3} - i}{2} = -\left(\frac{\sqrt{3} + i}{2}ight) = -z$.इसलिए,$\left(z^{101} + i^{103}ight)^{105} = (-z)^{105} = -z^{105}$.$z^{105} = \left(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}ight)^{105} = \cos \frac{105\pi}{6} + i \sin \frac{105\pi}{6} = \cos \frac{35\pi}{2} + i \sin \frac{35\pi}{2}$.चूंकि $\frac{35\pi}{2} = 18\pi - \frac{\pi}{2}$,इसलिए $z^{105} = \cos \left(-\frac{\pi}{2}ight) + i \sin \left(-\frac{\pi}{2}ight) = 0 - i = -i$.अंत में,$-z^{105} = -(-i) = i$.

यदि $z = \frac{\sqrt{3} + i}{2}$ है,तो $\left(z^{101} + i^{103}\right)^{105} = $

Similar Questions

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

यदि $n$ एक पूर्णांक है और $Z = \cos \theta + i \sin \theta$,जहाँ $\theta \neq (2n + 1) \frac{\pi}{2}$,तो $\frac{1 + Z^{2n}}{1 - Z^{2n}} = $

यदि ${\left( {\frac{{1 + i\sqrt 3 }}{{1 - i\sqrt 3 }}} \right)^n}$ एक पूर्णांक है,तो $n$ का न्यूनतम धनात्मक पूर्णांक मान क्या है?

माना $A_r = \left(x+\frac{1}{x}\right)^3 \cdot \left(x^2+\frac{1}{x^2}\right)^3 \cdot \left(x^3+\frac{1}{x^3}\right)^3 \cdots \left(x^r+\frac{1}{x^r}\right)^3$. यदि $x^2+x+1=0$ है,तो $\frac{1}{A_3}+\frac{1}{A_6}+\frac{1}{A_9}+\frac{1}{A_{12}}+\cdots \infty =$

$\sum_{k=1}^6\left(\sin \frac{2 \pi k}{7}-i \cos \frac{2 \pi k}{7}\right)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module