prove that ((i + ✓3)/2)^100 + ((i - ✓3)/2)^100 = -1
lhs is ((i + ✓3)/2)^100 + ((i - ✓3)/2)^100
Determine an expression for the 1st term ((i + ✓3)/2)^100
((i + ✓3)/2)^2 = (1/4)(-1 + 2i√3 + 3) = (1 + i√3)/2
((i + ✓3)/2)^4 = ((1 + i√3)/2)^2 = (1/4)(1 + 2i√3 – 3) = (-1 + i√3)/2
((i + ✓3)/2)^8 = ((-1 + i√3)/2)^2 = (1/4)(1 - 2i√3 – 3) = (-1 - i√3)/2
((i + ✓3)/2)^16 = ((-1 - i√3)/2)^2 = (1/4)(1 + 2i√3 – 3) = (-1 + i√3)/2
((i + ✓3)/2)^32 = ((-1 + i√3)/2)^2 = (1/4)(1 - 2i√3 – 3) = (-1 - i√3)/2
((i + ✓3)/2)^64 = ((-1 - i√3)/2)^2 = (1/4)(1 + 2i√3 – 3) = (-1 + i√3)/2
Now,
((i + ✓3)/2)^100 = ((i + ✓3)/2)^4 ((i + ✓3)/2)^32 ((i + ✓3)/2)^64
((i + ✓3)/2)^100 = (-1 + i√3)/2. (-1 - i√3)/2. (-1 + i√3)/2
((i + ✓3)/2)^100 = (-1 + i√3)/2.(1/4 – (-3)/4)
((i + ✓3)/2)^100 = (-1 + i√3)/2
Repeating for the 2nd term ((i – ✓3)/2)^100
((i – ✓3)/2)^2 = (1/4)(-1 – 2i√3 + 3) = (1 – i√3)/2
((i – ✓3)/2)^4 = ((1 – i√3)/2)^2 = (1/4)(1 – 2i√3 – 3) = (-1 – i√3)/2
((i – ✓3)/2)^8 = ((-1 – i√3)/2)^2 = (1/4)(1 + 2i√3 – 3) = (-1 + i√3)/2
((i – ✓3)/2)^16 = ((-1 + i√3)/2)^2 = (1/4)(1 – 2i√3 – 3) = (-1 – i√3)/2
((i – ✓3)/2)^32 = ((-1 – i√3)/2)^2 = (1/4)(1 + 2i√3 – 3) = (-1 + i√3)/2
((i – ✓3)/2)^64 = ((-1 + i√3)/2)^2 = (1/4)(1 – 2i√3 – 3) = (-1 – i√3)/2
Now,
((i – ✓3)/2)^100 = ((i – ✓3)/2)^4 ((i – ✓3)/2)^32 ((i – ✓3)/2)^64
((i – ✓3)/2)^100 = (-1 – i√3)/2. (-1 + i√3)/2. (-1 – i√3)/2
((i – ✓3)/2)^100 = (-1 – i√3)/2.(1/4 – (-3)/4)
((i – ✓3)/2)^100 = (-1 – i√3)/2
Adding together the two final results,
((i + ✓3)/2)^100 + ((i – ✓3)/2)^100 = (-1 + i√3)/2 + (-1 – i√3)/2 = -1/2 – 1/2
((i + ✓3)/2)^100 + ((i – ✓3)/2)^100 = -1 = rhs