Question

If \( z = \cos\theta + i \sin\theta \), then the value of \( z^{100} + \frac{1}{z^{100}} \) is

• A. \( 2 \cos 100\theta \)
• B. \( 2^{100} \cos\theta \)
• C. \( 2i \sin 100\theta \)
• D. \( 2^{100} \sin\theta \)

Hint:

For complex numbers in polar form \( z = e^{i\theta} \), powers of \( z \) result in a rotation by multiples of \( \theta \). Use the Euler's formula to simplify such expressions.

Answer 1

The Correct Option is A

Step 1: Since \( |z|=1 \), \( \dfrac{1}{z} = \bar z \), so \( z^{100}+\dfrac{1}{z^{100}} = z^{100}+\overline{z^{100}} = 2\,\mathrm{Re}(z^{100}) \).
Step 2: By De Moivre's theorem, \( z^{100}=\cos100\theta+i\sin100\theta \), so \( \mathrm{Re}(z^{100})=\cos100\theta \).
\[ \boxed{2\cos100\theta} \]

Answer 2

A third approach uses the recurrence relation satisfied by \( w_n = z^n + \dfrac{1}{z^n} \). Since \( w_1 = z+\dfrac1z = 2\cos\theta \), and multiplying by \( z+1/z \) gives the identity \( w_n\cdot w_1 = w_{n+1}+w_{n-1} \), i.e. \( w_{n+1} = 2\cos\theta\cdot w_n - w_{n-1} \), with \( w_0=2 \).

Solving this recurrence (whose characteristic roots are \( e^{i\theta}, e^{-i\theta} \)) gives the closed form \( w_n = 2\cos(n\theta) \) for every integer \( n\ge0 \), so in particular \( w_{100} = z^{100}+\dfrac{1}{z^{100}} = 2\cos(100\theta) \).

1. \( 2\cos100\theta \): Matches the closed-form solution of the recurrence for \( n=100 \) exactly.
2. \( 2^{100}\cos\theta \): The recurrence's closed form is a cosine of \(100\theta\), not \(2^{100}\) times \(\cos\theta\); this does not solve the recurrence and is incorrect.
3. \( 2i\sin100\theta \): This would come from the difference \(z^{100}-1/z^{100}\), which satisfies a different (sine) recurrence, not the sum considered here, so it is incorrect.
4. \( 2^{100}\sin\theta \): Does not solve the stated recurrence at all, so incorrect.

The recurrence relation for \( w_n \) confirms the same closed form as before.

Therefore, the correct answer is \( 2\cos100\theta \).