Question

Use De Moivre’s theorem to find (√3+i)^3 ​

Answer 1

Solution:

(By De Movire's Theorem)

$(Cos\: \theta \: + \: i\: sin\: \theta)^{n} \: = \: cos\: n\theta \: + \: i\: sin\: n\theta$

Let z = (√3 + i)

|z| = √ (a² + b²)

Now , a = √3 ; b = 1

|z| = √ [(√3)² + (1)²]

|z| = √ [(3) + 1 ]

|z| = √ [4]

|z| = 2

r = 2

$\theta \: = \: tan^{-1}\: \dfrac{b}{a}$

$\theta \: = \: tan^{-1}\: \dfrac{1}{\sqrt3}$

$\theta \: = \: \dfrac{\pi}{6}$

Polar form :

$z = 2\Big(cos\: \dfrac{\pi}{6} \:+ \:isin\:\dfrac{\pi}{6}\Big)$

Expansion

$[r(cos\: \theta \:+ \:isin\:\theta]^{n} \: = \: r^{n}[cos\:n\theta \: + \: isin\: n\theta]$

$\Big[2\Big(cos\: \dfrac{\pi}{3} \:+ \:isin\:\dfrac{\pi}{6}\Big)\Big]^{3} \: = \: 2^{3}\Big[cos\:3\dfrac{\pi}{6} \: + \: isin\: 3\dfrac{\pi}{6}\Big]$

$\Big[2\Big(cos\: \dfrac{\pi}{6} \:+ \:isin\:\dfrac{\pi}{6}\Big)\Big]^{3} \: =\: 8\Big[cos\: \dfrac{\pi }{2}\: + \: isin\: \dfrac{\pi}{2}\Big]$

$\Big[2\Big(cos\: \dfrac{\pi}{6} \:+ \:isin\:\dfrac{\pi}{6}\Big)\Big]^{3} \: =\:[8(cos \: 90^{\circ} + isin\: 90^{\circ})]$

Cos 90° = 0

Sin 90° = 1

Final Solution :

8(0 + i)

$\boxed{8i}$

Answer 2

Answer:

Step-by-step explanation:

Content:

De Moivre's theorem:-

(cosø+isinø)^n=(cos nø+isin nø)

We multiply and divide by 8,to bring to a form for applying the theorem.

cos(π/6)=✓3/2

sin(π/6)=1/2

cos(π/2)=0

sin(π/2)=1

We get a purely imaginary complex number 8i as the answer.