Prove that,
[tex]\bold{(\dfrac{i - \sqrt{3}}{i + \sqrt{3}})^{200} + (\dfrac{i +\sqrt{3}}{- i + \sqrt{3}})^{200}=-1}[/tex]
where i² = - 1
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Prove that,
[tex]\bold{(\dfrac{i - \sqrt{3}}{i + \sqrt{3}})^{200} + (\dfrac{i +\sqrt{3}}{- i + \sqrt{3}})^{200}=-1}[/tex]
where i² = - 1
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Answer:
[tex] \sf \: Now, \frac{i - \sqrt{3}}{i + \sqrt{3}} \\ \\ [/tex]
{ We rationalise the denominator by multiplying both the numerator and the denominator by (i - √3) }
[tex] \sf= \dfrac{(i-\sqrt{3})(i-\sqrt{3})}{(i + \sqrt{3})(i - \sqrt{3})}\\ \\ \sf= \dfrac{i^{2} - 2\sqrt{3}i + 3}{i^{2} - 3}\\ \\ \sf= \dfrac{- 1 - 2\sqrt{3}i + 3}{ - 1 - 3}[/tex]
, since i² = - 1
[tex] \sf= \dfrac{2 - 2\sqrt{3}i}{-4}\\ \\ \sf= - \dfrac{1}{2}+ i \dfrac{\sqrt{3}}{2}\\ \\ \sf= cos\dfrac{2\pi}{3} + i\:sin\dfrac{2\pi}{3}[/tex]
By De Moivre' theorem, we get
[tex] \sf \bigg(cos\dfrac{2\pi}{3} + i\:sin\dfrac{2\pi}{3} \bigg)^{200}\\ \\ \sf= cos\dfrac{400\pi}{3} + i\:sin\dfrac{400\pi}{3}[/tex]
[tex] \sf \: \to \small{\underline{{{ \bigg(\dfrac{i - \sqrt{3}}{i + \sqrt{3}}} \bigg)^{200}=cos\dfrac{2\pi}{3} + i\:sin\dfrac{4\pi}{3}}}}[/tex]
[tex] \sf \: Also, \dfrac{i + \sqrt{3}}{- i + \sqrt{3}}[/tex]
We rationalise the denominator by multiplying both the numerator and the denominator by (i + √3)
[tex]\\ \\ \sf= - \dfrac{(i + \sqrt{3}) (i + \sqrt{3})}{(i - \sqrt{3}) (i + \sqrt{3})}\\ \\ \sf= - \dfrac{i^{2} + 2\sqrt{3}i + 3}{i^{2} - 3} \\ \\ \sf= - \dfrac{- 1 + 2\sqrt{3}i + 3}{ - 1 - 3 }\\ \\ \sf= - \dfrac{2 + 2\sqrt{3}i}{-4}\sf \:= \dfrac{1}{2} + i \dfrac {\sqrt{3}}{2} \\ \\ \sf \: = cos\dfrac{\pi}{3}+ i\:sin\dfrac{\pi}{3}[/tex]
By De Moivre' theorem, we get
[tex] \sf \: \bigg(cos\dfrac{\pi}{3}+ i\:sin\dfrac{\pi}{3} \bigg)^{200} \\ \\ \sf \: = cos\dfrac{200\pi}{3}+ i\:sin\dfrac{200\pi}{3} \\ \\ \sf \: = cos\dfrac{2\pi}{3}+ i\:sin\dfrac{\pi}{3}[/tex]
[tex] \sf\to \small{\underline{{ \bigg(\dfrac{i +\sqrt{3}}{- i + \sqrt{3}} \bigg)^{200} = cos\dfrac{2\pi}{3}+ i\:sin\dfrac{\pi}{3}}}}[/tex]
Adding (1) and (2), we get
[tex] \sf \bigg(\dfrac{i - \sqrt{3}}{i + \sqrt{3}} \big)^{200} + \bigg(\dfrac{i +\sqrt{3}}{- i + \sqrt{3}} \bigg)^{200}\\ \\ \sf \: = \small{cos\dfrac{2\pi}{3} + i\:sin\dfrac{4\pi}{3} + cos\dfrac{2\pi}{3}+ i\:sin\dfrac{\pi}{3}} \\ \\ \sf \: = 2cos\dfrac{2\pi}{3} + i \: \bigg(- sin\dfrac{\pi}{3}+ sin{\pi}{3}\bigg)\\ \\ \sf= 2 cos\dfrac{2\pi}{3} \\ \\ \sf= 2 \times (-\dfrac{1}{2}) \\ \\ \red{= 1}[/tex]
Hence, proved.