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Use DeMoivre's theorem to evaluate the expression

[sqrt 3( cos 5pi/3 + i sin 5pi/3)]^4 ? write the answer in rectangular form


a. 9sqrt3/2 + 9/2 i

b. 9sqrt3/2 - 9/2 i

c. -9/2 - 9sqrt3/2 i

d. -9/2 + 9sqrt3/2 i

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Matheng
DeMoivre's theorem

if z = a ( cos θ + i sin θ)
∴ z^n = a^n ( cos nθ + i sin nθ)

For the given complex number ⇒⇒⇒ [ √3 ( cos 5π/3 + i sin 5π/3 ) ]⁴
[ √3 ( cos 5π/3 + i sin 5π/3 ) ]⁴ = (√3)⁴  ( cos 4*5π/3 + i sin 4*5π/3 )
= 9 ( cos 20π/3 + i sin 20π/3 )
= 9 ( cos 2π/3 + i sin 2π/3 )
 = 9 ( -1/2 + i √3 /2 )
= -9/2 + 9√3 /2 i

note: 20π/3 = 2π/3 + 6π = 2π/3 + 3 *2π = 2π/3

∴ The correct answer is option d
   d. -9/2 + 9sqrt3/2 i

Answer:

Option d - [tex]\sqrt3(\cos (\frac{5\pi}{3})+i\sin(\frac{5\pi}{3}))^4=-\frac{9}{2}+\frac{9\sqrt3}{2}i[/tex]      

Step-by-step explanation:

Given : [tex]\sqrt3(\cos (\frac{5\pi}{3})+i\sin(\frac{5\pi}{3}))^4[/tex]

To find : Use DeMoivre's theorem to evaluate the expression?

Solution :

DeMoivre's theorem state that, for complex number

If [tex]z = r(\cos\theta+ i\sin \theta)[/tex] then [tex]z^n = r^n(\cos n\theta+ i\sin n\theta)[/tex]

We have given,

[tex]\sqrt3(\cos (\frac{5\pi}{3})+i\sin(\frac{5\pi}{3}))^4[/tex]

On comparing [tex]r=\sqrt3[/tex] and n=4

Applying DeMoivre's theorem,

[tex]=(\sqrt3)^4(\cos 4(\frac{5\pi}{3})+i\sin4(\frac{5\pi}{3}))[/tex]

[tex]=9(\cos (\frac{20\pi}{3})+i\sin(\frac{20\pi}{3}))[/tex]

[tex]=9(\cos (6\pi+\frac{2\pi}{3})+i\sin(6\pi+\frac{2\pi}{3}))[/tex]

[tex]=9(\cos (\frac{2\pi}{3})+i\sin(\frac{2\pi}{3}))[/tex]

We know, the value of

[tex]\cos (\frac{2\pi}{3})=-\frac{1}{2},\sin (\frac{2\pi}{3})=\frac{\sqrt3}{2}[/tex]

[tex]=9(-\frac{1}{2}+i\frac{\sqrt3}{2})[/tex]

[tex]=-\frac{9}{2}+i\frac{9\sqrt3}{2}[/tex]    

Therefore, Option d is correct.

[tex]\sqrt3(\cos (\frac{5\pi}{3})+i\sin(\frac{5\pi}{3}))^4=-\frac{9}{2}+\frac{9\sqrt3}{2}i[/tex]      

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