Complex Numbers
Powers of complex numbers and geometric series
Grade 11
Question:
<p>Let \(z = 1 + ai\) where \(a > 0\). If \(z^3\) is a real number, the value of \(a\) is \(\sqrt{3}\). Then \(1 + z + z^2 + \cdots + z^{11}\) equals:</p>
<p>\(-1365\sqrt{3}\,i\)</p>
<p>\(1365\sqrt{3}\,i\)</p>
<p>\(-1365\)</p>
<p>\(1365\)</p>
Step-by-Step Solution
Key Concept: If z³ is real and z = 1 + ai, then Im(z³) = 0. Expand z³ using (1+ai)³ and set imaginary part to zero to find a, then use the geometric series formula for the sum.
<p><strong>Step 1: Find a using the condition that z³ is real.</strong></p><p>z = 1 + ai, so z³ = (1 + ai)³ = 1 + 3ai + 3a²i²(1) + a³i³</p><p>= 1 + 3ai - 3a² - a³i = (1 - 3a²) + i(3a - a³)</p><p>For z³ to be real: 3a - a³ = 0 ⟹ a(3 - a²) = 0</p><p>Since a > 0: a² = 3 ⟹ a = √3 ✓</p><p><strong>Step 2: Calculate z with a = √3.</strong></p><p>z = 1 + √3i, and z³ = (1 - 9) + i(3√3 - 3√3) = -8</p><p><strong>Step 3: Use geometric series formula.</strong></p><p>S = 1 + z + z² + ... + z¹¹ = (1 - z¹²)/(1 - z) = (1 - (z³)⁴)/(1 - z)</p><p>= (1 - (-8)⁴)/(1 - z) = (1 - 4096)/(1 - 1 - √3i) = -4095/(-√3i)</p><p>= 4095/(√3i) = 4095·(-i)/(√3·(-1)) = 4095i/√3 = 1365√3·i</p><p>Simplifying: = (1 + z + z² + ... + z¹¹) = 1365√3·i or equivalently (after rationalization) = <strong>-1365i/√3 · (-√3) = 1365i√3</strong></p><p>∴ Answer: A</p>
Correct Answer: A