Complex Numbers Question - None

Question

If α = \cos \frac{8π}{11} + i \sin \frac{8π}{11}, where i = \sqrt{-1}, then ext{Re}(α + α^2 + α^3 + α^4 + α^5) is

Accepted Answer

Step 1: Express α in exponential form.

Given: α = cos(8π/11) + i·sin(8π/11) = e^(i·8π/11)

Step 2: Verify that α is an 11th root of unity.

α^11 = e^(i·8π·11/11) = e^(i·8π) = 1

So α is an 11th root of unity (specifically, α = e^(i·8π/11)).

Step 3: Find the sum S = α + α^2 + α^3 + α^4 + α^5 using the geometric series formula.

S = α(1 - α^5)/(1 - α) [geometric series with first term α, ratio α, and 5 terms]

S = α(1 - α^5)/(1 - α) = (α - α^6)/(1 - α)

Step 4: Simplify using α^11 = 1.

Multiply numerator and denominator by (1 - α^11) approach, or directly compute:

S = (e^(i·8π/11) - e^(i·48π/11))/(1 - e^(i·8π/11))

Note: 48π/11 = 44π/11 + 4π/11 = 4π + 4π/11, so e^(i·48π/11) = e^(i·4π/11)

S = (e^(i·8π/11) - e^(i·4π/11))/(1 - e^(i·8π/11))

Step 5: Alternative approach - use that 1 + α + α^2 + ... + α^10 = 0 for the 11th root of unity.

From 1 + α + α^2 + ... + α^10 = 0, we get:

α + α^2 + α^3 + α^4 + α^5 = -(1 + α^6 + α^7 + α^8 + α^9 + α^10)

Step 6: Compute Re(α + α^2 + α^3 + α^4 + α^5) directly.

Re(α) = cos(8π/11), Re(α^2) = cos(16π/11), Re(α^3) = cos(24π/11), Re(α^4) = cos(32π/11), Re(α^5) = cos(40π/11)

Simplifying angles: 16π/11, 24π/11 = 2π + 2π/11, 32π/11 = 2π + 10π/11, 40π/11 = 4π - 4π/11

Re(S) = cos(8π/11) + cos(16π/11) + cos(2π/11) + cos(10π/11) + cos(4π/11)

Using symmetry and the relation for roots of unity: Re(α + α^2 + α^3 + α^4 + α^5) = -1/2

∴ Answer: B

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