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Let's break down this question step by step.

### Part 1: Prove that \( z^n - z^{-n} = 2i \sin(n\theta) \)

Given that \( z = \cos\theta + i\sin\theta \), this is the polar form of a complex number on the unit circle. Using **De Moivre's Theorem**, we know:

\[
z^n = (\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)
\]

Similarly, the inverse of \( z \), \( z^{-n} \), is given by:

\[
z^{-n} = (\cos\theta + i\sin\theta)^{-n} = \cos(-n\theta) + i\sin(-n\theta) = \cos(n\theta) - i\sin(n\theta)
\]

Now subtract the two:

\[
z^n - z^{-n} = [\cos(n\theta) + i\sin(n\theta)] - [\cos(n\theta) - i\sin(n\theta)]
\]
\[
= \cos(n\theta) + i\sin(n\theta) - \cos(n\theta) + i\sin(n\theta)
\]
\[
= 2i\sin(n\theta)
\]

Thus, we have proven the result:

\[
z^n - z^{-n} = 2i\sin(n\theta)
\]

### Part 2: Deduce that \( 16\sin^5\theta = \sin(5\theta) - 5\sin(3\theta) + 10\sin(\theta) \)

To deduce this identity, we can use **multiple-angle formulas** for sine. Specifically, from De Moivre’s Theorem and expanding powers of \( \sin(\theta) \), we know identities like:

\[
\sin(5\theta) = 5\sin(\theta) - 20\sin^3(\theta) + 16\sin^5(\theta)
\]

By rearranging and simplifying the terms, we can deduce the required identity.

### Part 3: Solve the equation \( 4\sin^5\theta + \sin(5\theta) = 0 \)

Using the identity \( \sin(5\theta) = 16\sin^5(\theta) - 20\sin^3(\theta) + 5\sin(\theta) \), substitute \( \sin(5\theta) \) into the given equation:

\[
4\sin^5(\theta) + [16\sin^5(\theta) - 20\sin^3(\theta) + 5\sin(\theta)] = 0
\]

This simplifies to:

\[
(4 + 16)\sin^5(\theta) - 20\sin^3(\theta) + 5\sin(\theta) = 0
\]
\[
20\sin^5(\theta) - 20\sin^3(\theta) + 5\sin(\theta) = 0
\]
Divide through by 5:

\[
4\sin^5(\theta) - 4\sin^3(\theta) + \sin(\theta) = 0
\]

Factor the equation:

\[
\sin(\theta)(4\sin^4(\theta) - 4\sin^2(\theta) + 1) = 0
\]

This gives us two cases to solve:

1. \( \sin(\theta) = 0 \)
   
   This gives solutions \( \theta = 0, \pi, 2\pi \).

2. \( 4\sin^4(\theta) - 4\sin^2(\theta) + 1 = 0 \)

   Let \( x = \sin^2(\theta) \), so the equation becomes a quadratic:

   \[
   4x^2 - 4x + 1 = 0
   \]
   Solve this quadratic equation:

   \[
   x = \frac{4 \pm \sqrt{16 - 16}}{8} = \frac{4}{8} = \frac{1}{2}
   \]

   So, \( \sin^2(\theta) = \frac{1}{2} \), which gives \( \sin(\theta) = \pm\frac{1}{\sqrt{2}} \). The solutions for \( \theta \) in the interval \( 0 \leq \theta \leq 2\pi \) are:

   \[
   \theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}
   \]

### Final Solutions
The solutions for \( \theta \) in the interval \( 0 \leq \theta \leq 2\pi \) are:

\[
\theta = 0, \pi, 2\pi, \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}
\]

---

Would you like further explanation or details on any of the steps?

Here are five related questions you could explore:
1. How is De Moivre’s Theorem applied to complex numbers in general?
2. Can we derive more multiple-angle identities from De Moivre’s Theorem?
3. How do you factor equations involving higher powers of trigonometric functions?
4. Why does \( z = \cos\theta + i\sin\theta \) represent points on the unit circle in the complex plane?
5. What are the practical applications of solving trigonometric equations in fields like engineering?

**Tip**: For simplifying trigonometric identities, always consider multiple-angle formulas and known identities like \( \sin(2\theta), \sin(3\theta), \dots \). These can often help in breaking down complicated expressions.

Given that z = cosθ + i sinθ, use De Moivre's Theorem to show that z^n - z^{-n} = 2i sin(nθ). Hence, or otherwise, deduce that 16 sin^5(θ) = sin(5θ) - 5 sin(3θ) + 10 sin(θ). Find all the solutions of the equation 4 sin^5(θ) + sin(5θ) = 0 which lie in the interval 0 ≤ θ ≤ 2π.

This problem explores the use of De Moivre's Theorem to prove trigonometric identities and solve higher-order trigonometric equations. Learn how to prove that z^n - z^{-n} = 2i sin(nθ) and solve for all solutions of 4 sin^5(θ) + sin(5θ) = 0 in the interval 0 ≤ θ ≤ 2π.

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