If $z = \cos\theta + i\sin\theta$, find an expression for $z^n + \dfrac{1}{z^n}$.
A$\sin 2n\theta$
B$\cos n\theta$
C$2\cos n\theta$
D$2i\sin n\theta$
Answer & Solution
Correct answer: C. $2\cos n\theta$
**Apply De Moivre.** $z^n = \cos n\theta + i\sin n\theta$.
**Compute the reciprocal.** Since $|z| = 1$, $\dfrac{1}{z^n} = \overline{z^n} = \cos n\theta - i\sin n\theta$ (conjugate of a unit-modulus complex number is its inverse).
**Sum.** $z^n + \dfrac{1}{z^n} = 2\cos n\theta$ — the imaginary parts cancel.
**Symmetric companion.** Subtracting instead gives $z^n - \dfrac{1}{z^n} = 2i\sin n\theta$ (option B) — useful for deriving multiple-angle formulae, but it's not what the question asks for.
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