Question

Prove that,

$\bold{(\dfrac{i - \sqrt{3}}{i + \sqrt{3}})^{200} + (\dfrac{i +\sqrt{3}}{- i + \sqrt{3}})^{200}=-1}$

where i² = - 1

Answer 1

$\boxed{\underline{\textsf{Method -- 1 ( De Moivre's ) :}}}$

Now, $\frac{i - \sqrt{3}}{i + \sqrt{3}}$

{ We rationalise the denominator by multiplying both the numerator and the denominator by (i - √3) }

= $\dfrac{(i-\sqrt{3})(i-\sqrt{3})}{(i + \sqrt{3})(i - \sqrt{3})}$

= $\dfrac{i^{2} - 2\sqrt{3}i + 3}{i^{2} - 3}$

= $\dfrac{- 1 - 2\sqrt{3}i + 3}{ - 1 - 3}$ , since i² = - 1

= $\dfrac{2 - 2\sqrt{3}i}{-4}$

= $- \dfrac{1}{2}+ i \dfrac{\sqrt{3}}{2}$

= $cos\dfrac{2\pi}{3} + i\:sin\dfrac{2\pi}{3}$

By De Moivre' theorem, we get

$(cos\dfrac{2\pi}{3} + i\:sin\dfrac{2\pi}{3})^{200}$

= $cos\dfrac{400\pi}{3} + i\:sin\dfrac{400\pi}{3}$

= $cos\dfrac{2\pi}{3} + i\:sin\dfrac{4\pi}{3}$

$\to \small{\boxed{{(\dfrac{i - \sqrt{3}}{i + \sqrt{3}}})^{200}=cos\dfrac{2\pi}{3} + i\:sin\dfrac{4\pi}{3}}}$
.....(1)

Also, $\dfrac{i + \sqrt{3}}{- i + \sqrt{3}}$

We rationalise the denominator by multiplying both the numerator and the denominator by (i + √3)

= $- \dfrac{(i + \sqrt{3}) (i + \sqrt{3})}{(i - \sqrt{3}) (i + \sqrt{3})}$

= $- \dfrac{i^{2} + 2\sqrt{3}i + 3}{i^{2} - 3}$

= $- \dfrac{- 1 + 2\sqrt{3}i + 3}{ - 1 - 3 }$

= $- \dfrac{2 + 2\sqrt{3}i}{-4}$

= $\dfrac{1}{2} + i \dfrac {\sqrt{3}}{2}$

= $cos\dfrac{\pi}{3}+ i\:sin\dfrac{\pi}{3}$

By De Moivre' theorem, we get

$(cos\dfrac{\pi}{3}+ i\:sin\dfrac{\pi}{3})^{200}$

= $cos\dfrac{200\pi}{3}+ i\:sin\dfrac{200\pi}{3}$

= $cos\dfrac{2\pi}{3}+ i\:sin\dfrac{\pi}{3}$

$\to \small{\boxed{(\dfrac{i +\sqrt{3}}{- i + \sqrt{3}})^{200} = cos\dfrac{2\pi}{3}+ i\:sin\dfrac{\pi}{3}}}$
.....(2)

Adding (1) and (2), we get

$(\dfrac{i - \sqrt{3}}{i + \sqrt{3}})^{200} + (\dfrac{i +\sqrt{3}}{- i + \sqrt{3}})^{200}$

$= \small{cos\dfrac{2\pi}{3} + i\:sin\dfrac{4\pi}{3} + cos\dfrac{2\pi}{3}+ i\:sin\dfrac{\pi}{3}}$

= $2cos\dfrac{2\pi}{3} + i (- sin\dfrac{\pi}{3}+ sin{\pi}{3})$

= $2 cos\dfrac{2\pi}{3}$

= $2*(-\dfrac{1}{2})$

= - 1

Hence, proved.

$\boxed{\underline{\textsf{Method -- 2 ( Cube roots of 1 ) :}}}$

Now, $\frac{i - \sqrt{3}}{i + \sqrt{3}}$

{ We rationalise the denominator by multiplying both the numerator and the denominator by (i - √3) }

$=\dfrac{(i-\sqrt{3})(i-\sqrt{3})}{(i + \sqrt{3})(i - \sqrt{3})}$

$=\dfrac{i^{2} - 2\sqrt{3}i + 3}{i^{2} - 3}$

$=\dfrac{- 1 - 2\sqrt{3}i + 3}{ - 1 - 3}$ , since i² = - 1

$=\dfrac{2 - 2\sqrt{3}i}{-4}$

$=\dfrac{-1+ i\sqrt{3}}{2}$

= ω, where ω is the imaginary cube root of 1

Also, $\dfrac{i + \sqrt{3}}{- i + \sqrt{3}}$

We rationalise the denominator by multiplying both the numerator and the denominator by (i + √3)

$=- \dfrac{(i + \sqrt{3}) (i + \sqrt{3})}{(i - \sqrt{3}) (i + \sqrt{3})}$

$=- \dfrac{i^{2} + 2\sqrt{3}i + 3}{i^{2} - 3}$

$=- \dfrac{- 1 + 2\sqrt{3}i + 3}{ - 1 - 3 }$

$=- \dfrac{2 + 2\sqrt{3}i}{-4}$

$=-\dfrac{-1- i \sqrt{3}}{2}$

= - ω², where ω² is the imaginary cube root of 1

Now, $(\dfrac{i - \sqrt{3}}{i + \sqrt{3}})^{200} + (\dfrac{i +\sqrt{3}}{- i + \sqrt{3}})^{200}$

= (ω)²⁰⁰ + (-ω²)²⁰⁰

= (ω³)⁶⁶ ω² + (ω³)¹³³ ω

= ω² + ω, since ω³ = 1

= - 1, since ω² + ω + 1 = 0

Hence, proved.

$\underline{\textsf{Finding cube roots of 1}}$

Let us take the following equation.

x³ = 1 .....(3)

⇒ x³ - 1 = 0

⇒ (x - 1) (x² + x + 1) = 0

Either x - 1 = 0 or, x² + x + 1 = 0

⇒ x = 1 , x² + x + 1 = 0

When, x² + x + 1 = 0, using Sridhar Acharya's formula, we get

x = $\dfrac{-1\pm \sqrt{1^{2}-(4*1*1)}}{2*1}$

= $\dfrac{-1\pm \sqrt{(1-4)}}{2}$

= $\dfrac{-1\pm \sqrt{(-3)}}{2}$

= $\bold{\dfrac{-1\pm i\sqrt{3}}{2}}$ , since i² = - 1

If we consider ω = $\dfrac{-1+ i\sqrt{3}}{2}$ , then it can be shown that

ω² = $\dfrac{-1-i\sqrt{3}}{2}$

Also ω being a root of (3) no. equation, we can write

ω² + ω + 1 = 0