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

z+1z=2cos⁡θ⇒z2−2zcos⁡θ+1=0⇒ z=cos⁡θ ± i sin⁡θNow, zn=cosnθ+isinnθ⇒1zn=cosnθ-isinnθ ∴zn+1zn=2cosnθ⇒z2n-2zncosnθ=-1

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

z+1z=2cos⁡θ⇒z2−2zcos⁡θ+1=0⇒ z=cos⁡θ ± i sin⁡θNow, zn=cosnθ+isinnθ⇒1zn=cosnθ-isinnθ ∴zn+1zn=2cosnθ⇒z2n-2zncosnθ=-1

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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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$\begin{array}{l} z + \frac{1}{z} = 2 \cos θ \Rightarrow z^{2} - 2 z \cos θ + 1 = 0 \\ \Rightarrow z = \cos θ \pm i \sin θ \end{array}$

Now,

$z^{n} = \cos n θ + i \sin n θ \Rightarrow \frac{1}{z^{n}} = \cos n θ - i \sin n θ ∴ z^{n} + \frac{1}{z^{n}} = 2 \cos n θ \Rightarrow z^{2 n} - 2 z^{n} \cos n θ = - 1$

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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