Complex Variables and Partial Differential Equations (3130005)

Question-1b

BE | Semester-3 Winter-2019 | 26-11-2019

Q1) (b)

State De-Movire’s formula and hence evaluate ${(1 + i \sqrt{3})}^{100} + {(1 - i \sqrt{3})}^{100} .$

De-Movire’s formula:

{(cosθ + i sinθ)}^{n} = \cos nθ + i \sin nθ; n \in ℚ - - - - - [1]

${(1 + i \sqrt{3})}^{100}$

Let,

z = 1 + i \sqrt{3}

\Rightarrow x = 1 & y = \sqrt{3}

r = \sqrt{x^{2} + y^{2}} = \sqrt{{(1)}^{2} + {(\sqrt{3})}^{2}} = \sqrt{1 + 3} = \sqrt{4} = 2

θ = \tan^{- 1} |\frac{y}{x}| = \tan^{- 1} |\frac{\sqrt{3}}{1}| = \tan^{- 1} \sqrt{3} = \frac{π}{3}

The polar form of complex number,

1 + i \sqrt{3} = r (cosθ + i sinθ) = 2 (\cos \frac{π}{3} + i \sin \frac{π}{3})

{(1 + i \sqrt{3})}^{100} = 2^{100} {(\cos \frac{π}{3} + i \sin \frac{π}{3})}^{100}

{(1 + i \sqrt{3})}^{100} = 2^{100} (\cos \frac{100 π}{3} + i \sin \frac{100 π}{3}) (∵ 1)

{(1 + i \sqrt{3})}^{100} = 2^{100} (- 0.5 + i \sin \frac{100 π}{3})

${(1 - i \sqrt{3})}^{100}$

Let,

z = 1 - i \sqrt{3}

\Rightarrow x = 1 & y = - \sqrt{3}

r = \sqrt{x^{2} + y^{2}} = \sqrt{{(1)}^{2} + {(- \sqrt{3})}^{2}} = \sqrt{1 + 3} = \sqrt{4} = 2

θ = \tan^{- 1} |\frac{y}{x}| = \tan^{- 1} |\frac{- \sqrt{3}}{1}| = - \tan^{- 1} \sqrt{3} = - \frac{π}{3}

The polar form of complex number,

1 - i \sqrt{3} = r (cosθ + i sinθ) = 2 (\cos \frac{π}{3} - i \sin \frac{π}{3})

{(1 - i \sqrt{3})}^{100} = 2^{100} {(\cos \frac{π}{3} - i \sin \frac{π}{3})}^{100}

{(1 + i \sqrt{3})}^{100} = 2^{100} (\cos \frac{100 π}{3} - i \sin \frac{100 π}{3}) (∵ 1)

{(1 + i \sqrt{3})}^{100} = 2^{100} (- 0.5 - i \sin \frac{100 π}{3})

{(1 + i \sqrt{3})}^{100} + {(1 - i \sqrt{3})}^{100}

= 2^{100} (- 0.5 + i \sin \frac{100 π}{3}) + 2^{100} (- 0.5 - i \sin \frac{100 π}{3})

= 2^{100} (- 0.5 + i \sin \frac{100 π}{3} - 0.5 - i \sin \frac{100 π}{3})

= 2^{100} (- 0.5 - 0.5)

= - 2^{100}

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