Use mathematical induction to prove De Moivre's theorem:
\((\cos \theta+i \,\sin \theta)^n=\cos n \theta+i \, \sin n \theta\)
for all integers \(n \geq 1\). (3 marks)
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\(\text{Proof (See Worked Solutions)}\)
\(\text{Prove} \ \ (\cos \theta+i \, \sin \theta)^n=\cos (n \theta)+i \sin (n \theta) \ \text {for} \ \ n \geq 1\)
\(\text{If} \ \ n=1:\)
\((\cos \theta+i \, \sin \theta)^1=\cos \theta+i \, \sin \theta=\cos (1 \theta)+i \, \sin (1 \theta)\)
\(\therefore \ \text{True for} \ \ n=1\)
\(\text{Assume true for} \ \ n=k:\)
\((\cos \theta+i \, \sin \theta)^k=\cos k \theta+i \,\sin k \theta\ \ldots\ (1)\)
\(\text{Prove true for} \ \ n=k+1:\)
\(\text{i.e.}\ \ (\cos \theta+i \,\sin \theta)^{k+1}=\cos (k+1) \theta+i \, \sin (k+1) \theta.\)
| \(\text{LHS}\) | \(=(\cos \theta+i \, \sin \theta)^{k+1}\) |
| \(=(\cos \theta+i \, \sin \theta)^k(\cos \theta+i \, \sin \theta)\) | |
| \(=(\cos k \theta+i \, \sin k \theta)(\cos \theta+i \, \sin \theta) \ \ \text{(using}\ (1)\ \text{above)}\) | |
| \(=\cos k \theta \, \cos \theta+i \, \cos k \theta \,\sin \theta+i \, \sin k \theta \,\cos \theta+i^2 \, \sin k \theta \,\sin \theta\) | |
| \(=(\cos k \theta \, \cos \theta-\sin k \theta \,\sin \theta)+i(\sin k \theta \,\cos \theta+\cos k \theta \,\sin \theta)\) | |
| \(=\cos (k+1) \theta+i \, \sin (k+1) \theta\) |
\(\Rightarrow \text{True for} \ \ n=k+1\)
\(\therefore \ \text{Since true for \(n=1\), by PMI, true for integers \(n \geqslant 1\).}\)