If z+(1/z)=2 cos θ, z ∈ C then z2 n-2 zn cos (n θ) is equal to

Question

If z+(1/z)=2 cos θ, z ∈ C then z2 n-2 zn cos (n θ) is equal to (A) 1 (B) 0 (C) -1 (D) -n

Tardigrade Admin · Accepted Answer

z+(1/z)=2 cos θ ⇒ z2-2 z cos θ+1=0 ⇒ z= cos θ ± i sin θ text Now, z2 n-2 zn cos (n θ) =zn[zn-2 cos (n θ)] =zn[ cos (n θ) ± i sin (n θ)-2 cos (n θ)] =-zn barzn=-1

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

1

0

-1

-n

$z + \frac{1}{z} = 2 cos θ \Rightarrow z^{2} - 2 z cos θ + 1 = 0$
$\Rightarrow z = cos θ \pm i sin θ$
$Now, z^{2 n} - 2 z^{n} cos (n θ)$
$= z^{n} [z^{n} - 2 cos (n θ)]$
$= z^{n} [cos (n θ) \pm i sin (n θ) - 2 cos (n θ)]$
$= - z^{n} \overset{z}{ˉ}^{n} = - 1$

Q. If z+z1​=2cosθ,z∈C then z2n−2zncos(nθ) is equal to

1

0

-1

-n

z+z1​=2cosθ⇒z2−2zcosθ+1=0 ⇒z=cosθ±isinθ Now, z2n−2zncos(nθ) =zn[zn−2cos(nθ)] =zn[cos(nθ)±isin(nθ)−2cos(nθ)] =−znzˉn=−1

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

$z + \frac{1}{z} = 2 cos θ \Rightarrow z^{2} - 2 z cos θ + 1 = 0$
$\Rightarrow z = cos θ \pm i sin θ$
$Now, z^{2 n} - 2 z^{n} cos (n θ)$
$= z^{n} [z^{n} - 2 cos (n θ)]$
$= z^{n} [cos (n θ) \pm i sin (n θ) - 2 cos (n θ)]$
$= - z^{n} \overset{z}{ˉ}^{n} = - 1$