Two values of (-8-8 sqrt{3} i)^{1/4} are | vedclass

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

Question

(A) Let $z = -8-8 \sqrt{3} i$. We can write this in polar form as $z = r(\cos 	heta + i \sin 	heta)$.Here,$r = \sqrt{(-8)^2 + (-8 \sqrt{3})^2} = \sq

Vedclass · Accepted Answer

$\sqrt{3}-i, -1-\sqrt{3} i$

Vedclass · Answer

(A) Let $z = -8-8 \sqrt{3} i$. We can write this in polar form as $z = r(\cos 	heta + i \sin 	heta)$.Here,$r = \sqrt{(-8)^2 + (-8 \sqrt{3})^2} = \sqrt{64 + 192} = \sqrt{256} = 16$.Since both real and imaginary parts are negative,the angle $	heta$ lies in the third quadrant.$	an 	heta = \frac{-8 \sqrt{3}}{-8} = \sqrt{3}$,so $	heta = \pi + \frac{\pi}{3} = \frac{4 \pi}{3}$.Thus,$z = 16(\cos(\frac{4 \pi}{3} + 2k \pi) + i \sin(\frac{4 \pi}{3} + 2k \pi))$ for $k = 0, 1, 2, 3$.The roots are given by $z^{1/4} = 16^{1/4}(\cos(\frac{4 \pi/3 + 2k \pi}{4}) + i \sin(\frac{4 \pi/3 + 2k \pi}{4})) = 2(\cos(\frac{\pi}{3} + \frac{k \pi}{2}) + i \sin(\frac{\pi}{3} + \frac{k \pi}{2}))$.For $k=0$: $2(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}) = 2(\frac{1}{2} + i \frac{\sqrt{3}}{2}) = 1 + i \sqrt{3}$.For $k=1$: $2(\cos \frac{5 \pi}{6} + i \sin \frac{5 \pi}{6}) = 2(-\frac{\sqrt{3}}{2} + i \frac{1}{2}) = -\sqrt{3} + i$.For $k=2$: $2(\cos \frac{4 \pi}{3} + i \sin \frac{4 \pi}{3}) = 2(-\frac{1}{2} - i \frac{\sqrt{3}}{2}) = -1 - i \sqrt{3}$.For $k=3$: $2(\cos \frac{11 \pi}{6} + i \sin \frac{11 \pi}{6}) = 2(\frac{\sqrt{3}}{2} - i \frac{1}{2}) = \sqrt{3} - i$.The values are $\sqrt{3}-i$ and $-\sqrt{3}+i$ (among others). Comparing with options,option $C$ contains $-\sqrt{3}+i$ and $\sqrt{3}+i$ is not a root,but checking the provided options,the correct set is $A$.

Two values of $(-8-8 \sqrt{3} i)^{1/4}$ are

Similar Questions

The value of $\frac{a + b\omega + c\omega^2}{b + c\omega + a\omega^2} + \frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2}$ is

The product of the distinct $(2n)^{\text{th}}$ roots of $1+i\sqrt{3}$ is equal to:

If $2+2 \sqrt{3} i=k(\cos \theta+i \sin \theta)$ where $k > 0$,then find the value of $\frac{1}{\sqrt{3}}[\cos 6 \theta+i \sin 6 \theta]$.

If $\alpha$ is a non-real root of $x^7=1$,then $\alpha(1+\alpha)(1+\alpha^2+\alpha^4) = $

If $x = \cos \theta + i\sin \theta$ and $y = \cos \phi + i\sin \phi$,then ${x^m}{y^n} + {x^{-m}}{y^{-n}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module