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  • Please help! - Complex numbers and quadratic equations - JEE Main-3

Let a=\cos \alpha +i\sin \alpha ,\: b=\cos \beta +i\sin \beta ,\: c=\cos \gamma +i\sin \gamma and \frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}=1  then \cos \left ( 2\alpha -2\beta \right )+\cos \left ( 2\beta -2\gamma \right )+\cos \left ( 2\gamma -2\alpha \right ) equals

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a^{2}=\cos 2\alpha +i\sin 2\alpha =e^{i2\alpha }

b^{2}=\cos 2\beta +i\sin 2\beta =e^{i2\beta }

c^{2}=\cos 2\gamma +i\sin 2\gamma =e^{i2\gamma }

\Rightarrow \frac{e^{i2\gamma }}{e^{i2\beta }}+\frac{e^{i2\beta }}{e^{i2\gamma }}+\frac{e^{i2\gamma }}{e^{i2\alpha }}=1

\Rightarrow \cos \left ( 2\alpha -2\beta \right )+\cos \left ( 2\beta -2\gamma \right )+\cos \left ( 2\gamma -2\alpha \right )=1\; \; and\; \; \sin \left ( 2\alpha -2\beta \right )+\sin \left ( 2\beta -2\gamma \right )+\sin \left ( 2\gamma -2\alpha \right )=0

De Moivre's Theorem -

\left ( \cos \Theta +i\sin \Theta \right )^{n}=\cos n\Theta +i\sin n\Theta \; \forall \; n\, \epsilon I

If n\, \epsilon \, Q , n\, \neq \, I, n=\frac{p}{q},  HCF\left ( p,q \right )=1 then \left ( \cos \Theta +i\sin \Theta \right )^{n} has \left | q \right | number of results, out of which one result will be \cos n\Theta +i\sin n \Theta.

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