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If $z + \frac{1}{z} = 2 \cos θ, z \in C$ then $z^{2 n} - 2 z^{n} \cos (n θ)$ is equal to

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Correct option is C

z+1z=2cos⁡θ⇒z2−2zcos⁡θ+1=0⇒ z=cos⁡θ ± i sin⁡θNow, z2n−2zncos⁡(nθ)=znzn−2cos⁡(nθ)=zn[cos⁡(nθ)±isin⁡(nθ)−2cos⁡(nθ)]=−znz¯n=−1

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If z+1z=2cos⁡θ,z∈C then z2n−2zncos⁡(nθ) is equal to

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If $z + \frac{1}{z} = 2 \cos θ, z \in C$ then $z^{2 n} - 2 z^{n} \cos (n θ)$ is equal to