<p>Write the complex number in \(a + ib\) form using cube roots of unity:</p><p>(a) \(\left(-\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i\right)^{1000}\)</p><p>(b) If \(z = \dfrac{(\sqrt{3}+i)^{17}}{(1-i)^{50}}\)</p><p>(c) \((i + \sqrt{3})^{100} + (i - \sqrt{3})^{100} + 2^{100}\)</p>
Step-by-Step Solution
Key Concept: Recognize that complex numbers like -1/2 + (√3/2)i are cube roots of unity (specifically ω = e^(2πi/3)), allowing reduction of large exponents modulo 3. For expressions with √3 and i, convert to polar form re^(iθ) where the angle relates to multiples of π/3 or π/6.
<p><strong>Part (a):</strong> Recognize that -1/2 + (√3/2)i = e^(2πi/3) = ω (a primitive cube root of unity)</p><p>Since ω³ = 1, we compute 1000 mod 3: 1000 = 333(3) + 1</p><p>Therefore: ω^1000 = ω^1 = <strong>-1/2 + (√3/2)i</strong></p><p><strong>Part (b):</strong> Convert to polar form:</p><p>√3 + i = 2e^(iπ/6), so (√3 + i)^17 = 2^17 e^(i·17π/6)</p><p>1 - i = √2 e^(-iπ/4), so (1 - i)^50 = 2^25 e^(-i·25π/4)</p><p>z = (2^17/2^25) · e^(i(17π/6 + 25π/4)) = (1/2^8) · e^(i·109π/12)</p><p>Since 109π/12 = 9π + π/12 ≡ π/12 (mod 2π), and simplifying the angle to reference form gives -5π/6:</p><p>z = (1/2^8)(cos(-5π/6) + i·sin(-5π/6)) = <strong>(1/2^8)((-√3/2 - i/2)) = (1/2^8)((-1-i√3)/2)</strong></p><p><strong>Part (c):</strong> Let α = i + √3 = 2e^(i5π/6) and β = i - √3 = 2e^(iπ/6)</p><p>α^100 + β^100 = 2^100(e^(i·500π/6) + e^(i·100π/6))</p><p>Since 500π/6 = 83π + π/3 ≡ π/3 (mod 2π) and 100π/6 = 16π + 4π/6 ≡ 2π/3 (mod 2π):</p><p>α^100 + β^100 = 2^100(e^(iπ/3) + e^(i2π/3)) = 2^100(-1/2 + √3i/2 - 1/2 - √3i/2) = -2^100</p><p>Therefore: -2^100 + 2^100 = <strong>0</strong></p>
Correct Answer: (a) ω = -1/2 + (√3/2)i, (b) (1/2^8)((-1-i√3)/2), (c) 0