Complex Numbers
Powers of complex numbers
Grade 11
Question:
<p>If \(z = \dfrac{\sqrt{3}}{2} + \dfrac{i}{2}\) \((i = \sqrt{-1})\), then \((1 + iz + z^5 + iz^8)^9\) is equal to __________.</p>
Step-by-Step Solution
Key Concept: Recognize that z = e^(iπ/6) is a 12th root of unity (z^12 = 1), allowing cyclic reduction of powers before expanding the bracket.
<p><strong>Step 1:</strong> Express z in exponential form. z = √3/2 + i/2 = cos(π/6) + i·sin(π/6) = e^(iπ/6)</p><p><strong>Step 2:</strong> Find the period. z^12 = e^(i·2π) = 1, so z is a 12th root of unity.</p><p><strong>Step 3:</strong> Reduce powers using z^12 = 1:<br/>• z^5 = e^(i5π/6)<br/>• z^8 = e^(i4π/3) = z^(-4) (since z^8 · z^4 = z^12 = 1)<br/>• iz = e^(iπ/2) · e^(iπ/6) = e^(i2π/3)</p><p><strong>Step 4:</strong> Simplify the bracket:<br/>1 + iz + z^5 + iz^8 = 1 + e^(i2π/3) + e^(i5π/6) + i·e^(i4π/3)<br/>= 1 + (-1/2 + i√3/2) + (-√3/2 + i/2) + i(-1/2 - i√3/2)<br/>= 1 + (-1/2 + i√3/2) + (-√3/2 + i/2) + (-i/2 + √3/2)<br/>= (1 - 1/2 - √3/2 + √3/2) + i(√3/2 + 1/2 - 1/2)<br/>= 1/2 + i√3/2 = e^(iπ/3)</p><p><strong>Step 5:</strong> Raise to the 9th power:<br/>(e^(iπ/3))^9 = e^(i3π) = e^(iπ) = -1</p><p>∴ Answer: <strong>-1</strong></p>
Correct Answer: -1