Question

if w=cos(2pie/3)-isin(4pie/3) find w5 without using a calculator​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\omega=\cos\left(\dfrac{2\pi}{3}\right)-i\,\sin\left(\dfrac{4\pi}{3}\right)$

$\implies\omega=\cos\left(\pi-\dfrac{\pi}{3}\right)-i\,\sin\left(\pi+\dfrac{\pi}{3}\right)$

$\implies\omega=-\cos\left(\dfrac{\pi}{3}\right)+i\,\sin\left(\dfrac{\pi}{3}\right)$

$\implies\omega=\cos\left(\pi-\dfrac{\pi}{3}\right)+i\,\sin\left(\pi-\dfrac{\pi}{3}\right)$

$\implies\omega=\cos\left(\dfrac{2\pi}{3}\right)+i\,\sin\left(\dfrac{2\pi}{3}\right)$

$\implies\omega=e^{\displaystyle\,i\,\dfrac{2\pi}{3}}$

$\implies\omega^5=e^{\displaystyle\,i\,\dfrac{2\pi}{3}\times5}$

$\implies\omega^5=e^{\displaystyle\,i\,\dfrac{10\pi}{3}}$

$\implies\omega^5=\cos\left(\dfrac{10\pi}{3}\right)+i\,\sin\left(\dfrac{10\pi}{3}\right)$

$\implies\omega^5=\cos\left(3\pi+\dfrac{\pi}{3}\right)+i\,\sin\left(3\pi+\dfrac{\pi}{3}\right)$

$\implies\omega^5=-\cos\left(\dfrac{\pi}{3}\right)-i\,\sin\left(\dfrac{\pi}{3}\right)$

$\implies\omega^5=-\dfrac{1}{2}-i\,\dfrac{\sqrt{3}}{2}$

$\implies\omega^5=-\dfrac{1}{2}\big(1+i\sqrt{3}\big)$

Answer 2

w⁵ = (-1/2) (1 + i√3) if w = cos (2π/3) - i sin( 4π/3)

Given : w = cos (2π/3) - i sin( 4π/3)

To Find : w⁵

Solution:

w = cos (2π/3) - i sin( 4π/3)

=> w = cos (2π/3) - i sin( 2π - 2π/3)

sin (2π - x) = - sin  x

=> w = cos (2π/3) + i sin(2π/3)

Using De Moivre's formula

(cosθ  + isinθ)ⁿ = cos(nθ) + i sin(nθ)

w⁵ = cos (5*2π/3) + i sin(5*2π/3)

w⁵ = cos (10π/3) + i sin(10π/3)

w⁵ = cos (2π+ 4π/3) + i sin(2π+ 4π/3)

=> w⁵ = cos (4π/3) + i sin (4π/3)

=>  w⁵ = -1/2  - i√3/2

=> w⁵ = (-1/2) (1 + i√3)

learn more:

What is the justification for step B? Review the proof of de Moivre's ...

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