Question

If alpha=e^i 8 pi/11then find re(a+a^2+a^3+a^4+a^5)

Answer 1

Answer:

Step-by-step explanation:

Given that a=e^((8i π)/11), then a^2=e^((16i π)/11).  

Similarly we have, a^3=e^((24i π)/11), a^4=e^((32i π)/11), and a^5=e^((40i π)/11).

Now, Re(a+a^2+a^3+a^4+a^5 )=Re(a)+Re(a^2 )+Re(a^3 )+Re(a^4 )+Re(a^5 ) and

Re(a)+Re(a^2 )+Re(a^3 )+Re(a^4 )+Re(a^5 )=1+1+1+1+1=5

Answer 2

Answer:

If alpha=e^i 8 pi/11 then  re(a+a^2+a^3+a^4+a^5) = -0.5  (-1/2)

Step-by-step explanation:

α = $e^{(\frac{\iota 8\pi}{11})}$ = cos (8π/11) + i Sin(8π/11)

α² = $e^{(\frac{\iota 16\pi}{11})}$ = cos (16π/11) + i Sin(16π/11)

α³ = $e^{(\frac{\iota 24\pi}{11})}$ = cos (24π/11) + i Sin(24π/11)

α⁴ = $e^{(\frac{\iota 32\pi}{11})}$ = cos (32π/11) + i Sin(32π/11)

α⁵ = $e^{(\frac{\iota 40\pi}{11})}$ = cos (40π/11) + i Sin(40π/11)

we know that

$e^{\iota \theta} = Cos\theta + \iota Sin\theta$

Re = Cosθ   Sinθ = Imaginary

Re(α + α² + α³ + α⁴ + α⁵) = cos (8π/11) + Cos(16π/11) + cos(24π/11) + Cos(32π/11) + Cos(40π/11)

Using Cosθ = Cos(θ±2π)   & Cosθ=cos(-θ)

= cos (8π/11)  + cos (6π/11)  + Cos(2π/11) + Cos(10π/11) + Cos(4π/11)

= -0.655  -0.142 + 0.841 - 0.96 + 0.416

= - 0.5

= -1/2