Problem Statement
Use the principle of mathematical induction to prove that for all integers $n \ge 1$:
$$(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$$
Verified Solution & Marking Scheme
Base Case n = 1
For $n = 1$:
$$\text{LHS} = (\cos\theta + i\sin\theta)^1 = \cos\theta + i\sin\theta$$
$$\text{RHS} = \cos(1\cdot\theta) + i\sin(1\cdot\theta) = \cos\theta + i\sin\theta$$
Since $\text{LHS} = \text{RHS}$, the statement holds true for $n = 1$.
Inductive Hypothesis
Assume the statement is true for $n = k$, where $k \in \mathbb{Z}^+$:
$$(\cos\theta + i\sin\theta)^k = \cos(k\theta) + i\sin(k\theta)$$
Inductive Step: Show statement holds for n = k + 1
Consider $n = k + 1$:
$$(\cos\theta + i\sin\theta)^{k+1} = (\cos\theta + i\sin\theta)^k (\cos\theta + i\sin\theta)$$
By induction hypothesis:
$$= [\cos(k\theta) + i\sin(k\theta)] [\cos\theta + i\sin\theta]$$
$$= \cos(k\theta)\cos\theta + i\cos(k\theta)\sin\theta + i\sin(k\theta)\cos\theta + i^2 \sin(k\theta)\sin\theta$$
$$= [\cos(k\theta)\cos\theta - \sin(k\theta)\sin\theta] + i[\sin(k\theta)\cos\theta + \cos(k\theta)\sin\theta]$$
Using compound angle formulae:
$$= \cos(k\theta + \theta) + i\sin(k\theta + \theta) = \cos((k+1)\theta) + i\sin((k+1)\theta)$$
Conclusion
Since the statement is true for $n = 1$, and its truth for $n = k$ implies its truth for $n = k + 1$, by the principle of mathematical induction, the statement is true for all $n \in \mathbb{Z}^+$.