<p>Express the following in <em>a + ib</em> form:<br>(b) \(\left(\dfrac{1+\cos\phi + i\sin\phi}{1+\cos\phi - i\sin\phi}\right)^n\)</p>
Step-by-Step Solution
Key Concept: Recognize the numerator and denominator as complex numbers in the form 1 + e^(iφ), then simplify using the half-angle identity: 1 + cos φ + i sin φ = 2cos(φ/2)e^(iφ/2). De Moivre's theorem then directly gives the result.
<p><strong>Step 1:</strong> Express numerator and denominator in exponential form.</p><p>Numerator: 1 + cos φ + i sin φ = (1 + cos φ) + i sin φ</p><p>Using the identity: 1 + cos φ = 2cos²(φ/2) and sin φ = 2sin(φ/2)cos(φ/2)</p><p>= 2cos(φ/2)[cos(φ/2) + i sin(φ/2)] = 2cos(φ/2)·e^(iφ/2)</p><p><strong>Step 2:</strong> Similarly for denominator:</p><p>1 + cos φ - i sin φ = 2cos(φ/2)[cos(φ/2) - i sin(φ/2)] = 2cos(φ/2)·e^(-iφ/2)</p><p><strong>Step 3:</strong> Form the ratio:</p><p>$$\frac{1+\cos\phi + i\sin\phi}{1+\cos\phi - i\sin\phi} = \frac{2\cos(\phi/2)e^{i\phi/2}}{2\cos(\phi/2)e^{-i\phi/2}} = e^{i\phi}$$</p><p><strong>Step 4:</strong> Apply De Moivre's theorem:</p><p>$$(e^{i\phi})^n = e^{in\phi} = \cos n\phi + i \sin n\phi$$<p><strong>∴ Answer: cos nφ + i sin nφ</strong></p>
Correct Answer: cos nφ + i sin nφ