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: Use Euler's formula.
We are given that \( z = \cos\theta + i \sin\theta \), which is the standard form of Euler’s formula. This means that \( z = e^{i\theta} \).

Step 2: Simplify the expression.
To find \( z^{100} + \frac{1}{z^{100}} \), we use the following property of complex exponentials: \[ z^{100} = e^{i100\theta} \quad \text{and} \quad \frac{1}{z^{100}} = e^{-i100\theta}. \] Thus, we have: \[ z^{100} + \frac{1}{z^{100}} = e^{i100\theta} + e^{-i100\theta} = 2 \cos(100\theta). \]
Step 3: Conclusion.
Thus, the value of \( z^{100} + \frac{1}{z^{100}} \) is \( 2 \cos 100\theta \), which corresponds to option (A).

Answer 2

Instead of Euler's formula, let's use the conjugate property directly: since \( |z| = |\cos\theta+i\sin\theta| = 1 \), we have \( \dfrac{1}{z} = \bar z \). This lets us rewrite \( z^{100} + \dfrac{1}{z^{100}} \) as \( z^{100} + \overline{z^{100}} \), which is twice the real part of \( z^{100} \).

By De Moivre's theorem, \( z^{100} = \cos(100\theta) + i\sin(100\theta) \), so its real part is \( \cos(100\theta) \). Therefore, \[ z^{100} + \overline{z^{100}} = 2\,\mathrm{Re}(z^{100}) = 2\cos(100\theta). \]

1. \( 2\cos100\theta \): This matches \( 2\,\mathrm{Re}(z^{100}) \) computed above exactly.
2. \( 2^{100}\cos\theta \): This would arise from confusing \( z^{100} \) (a rotation by \(100\theta\)) with \( (2\cos\theta)^{100} \)-type expressions; it does not follow from De Moivre's theorem correctly applied here, so it is incorrect.
3. \( 2i\sin100\theta \): This is \( z^{100} - \overline{z^{100}} \) (twice the imaginary part times \(i\)), not the sum \( z^{100}+\overline{z^{100}} \); it corresponds to a different combination and is incorrect.
4. \( 2^{100}\sin\theta \): Like option (B), this does not arise from any correct application of De Moivre's theorem to this sum, so it is incorrect.

Rewriting the reciprocal as a conjugate and taking twice the real part reproduces the same result as the exponential-form derivation.

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