Complex Numbers
Roots of Unity
Grade 11
Question:
<p>Find roots of the equation \((z + 1)^5 = (z - 1)^5\).</p>
<p>\(-i\cot\left(\dfrac{k\pi}{5}\right);\, k = 1, 2, 3, 4\)</p>
<p>\(i\cot\left(\dfrac{k\pi}{5}\right);\, k = 1, 2, 3, 4\)</p>
<p>\(-i\tan\left(\dfrac{k\pi}{5}\right);\, k = 1, 2, 3, 4\)</p>
<p>\(i\tan\left(\dfrac{k\pi}{5}\right);\, k = 1, 2, 3, 4\)</p>
Step-by-Step Solution
Key Concept: Rewrite the equation as [(z+1)/(z-1)]^5 = 1 to find the fifth roots of unity, then solve for z. The constraint z ≠ 1 eliminates one root.
<p><strong>Step 1:</strong> Divide both sides by (z-1)^5 (valid since z ≠ 1):</p><p>[(z+1)/(z-1)]^5 = 1</p><p><strong>Step 2:</strong> The fifth roots of unity are ω_k = e^(2πik/5) for k = 0, 1, 2, 3, 4.</p><p><strong>Step 3:</strong> Set (z+1)/(z-1) = ω_k, so z+1 = ω_k(z-1)</p><p><strong>Step 4:</strong> Solve for z: z+1 = ω_k·z - ω_k → z(1-ω_k) = -ω_k - 1 → z = (ω_k+1)/(ω_k-1)</p><p><strong>Step 5:</strong> When k = 0, ω_0 = 1, which gives 0/0 (undefined). This root must be excluded.</p><p><strong>Step 6:</strong> For k = 1, 2, 3, 4, we get four roots: z = (e^(2πik/5)+1)/(e^(2πik/5)-1) = cot(πk/5)·i</p><p>Equivalently: <strong>z = i·cot(πk/5) for k = 1, 2, 3, 4</strong></p><p>Or in exponential form: <strong>z = (1+e^(2πik/5))/(1-e^(2πik/5)) for k = 1, 2, 3, 4</strong></p><p>∴ Answer: A</p>
Correct Answer: A