prove that ((i+✓3)/2)^100+((i-✓3)/2)^100=-1

Question

Help me I want solve this problem currently

Answer 1

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

 

Answer 2

The equation x²-i-1=0 has the roots z₁=½(i+√3) and z₂=½(i-√3), which happen to be the two expressions given in the question. So z₁z₂=-1 (constant term) and z₁+z₂=i, the negative of the x coefficient.

-1=(z₁+z₂)²=z₁²+z₂²+2z₁z₂=z₁²+z₂²-2, so z₁²+z₂²=1.

Also, (z₁²+z₂²)²=1=z₁⁴+z₂⁴+2(z₁z₂)²=z₁⁴+z₂⁴+2, and z₁⁴+z₂⁴=1-2=-1.

z₁⁴+z₂⁴=-1.

Continuing, (z₁⁴+z₂⁴)²=1=z₁⁸+z₂⁸+2(z₁z₂)⁴=z₁⁸+z₂⁸+2, so z₁⁸+z₂⁸=-1.

(z₁⁸+z₂⁸)²=1=z₁¹⁶+z₂¹⁶+2(z₁z₂)⁸=z₁¹⁶+z₂¹⁶+2, so z₁¹⁶+z₂¹⁶=-1.

Thus, for exponents of the z’s equal to powers of 2, the sum will always be -1.

What is z₁³+z₂³?

-i=(z₁+z₂)³=z₁³+3z₁²z₂+3z₁z₂²+z₂³=z₁³+z₂³+3z₁z₂(z₁+z₂)=z₁³+z₂³-3i

z₁³+z₂³=2i.

We can find z₁⁵+z₂⁵=i similarly. 

If we tabulate the results we get:

n, 12m+n	Sum
1, 13, 25, ... 12m+1	i
2, 14, ... 12m+2 etc	1
3	2i
4, 16, 28, ..., 88, 100, ...	-1
5	i
6	-2
7	-i
8, 20, 32, ...	-1
9	-2i
10	1
11	-i
12, 12m	2

A pattern emerges: it repeats in groups of 12, so since 100=8×12+4, we need to look up n=4 to find out the result. So since this gives us -1, then when n=100 the sum is -1 QED. 

The reason for the pattern also arises from de Moivre, e^(nix) where x=π/6 or 30º and 12 of these make 2π or 360º. We can write z₁=e^(niπ/6) and z₂=-e^(-niπ/6).

Also, all powers of 2 can be expressed as a multiple of 3 plus 1 or 2, which is the reason why n=4 or 8 has the sum -1: 2ⁿ=12p+4q or 2ⁿ⁻²=3p+q. (All powers of 2 are 1 more or 1 less than a multiple of 3, which is self-evident.) 

Comments

Nice answer Rod. Probably more likely to be the expected solution than mine!

---

I wasn’t sure what mathematical method was expected to be included in the solution. Was it pure complex algebra, de Moivre, Argand diagrams, or what?

De Moivre seemed a possibility: e^(nix)=(cos(x)+isin(x))ⁿ=cos(nx)+isin(nx) where x=π/6 and this led to the following pretty diagrams:

The top diagram is an Argand diagram showing each of the points P for values of n from 1 to 12. The second diagram shows the progression of the sums S, z₁ⁿ+z₂ⁿ.

---

Yes, those expressions that you had, z1 and z2, looked so familiar. I thought complex algebra etc, but I was going to have to do some revision first before I could properly tackle the question. So I tried a simpler method first of all by squaring the terms z1 and z2, and noticed I could eliminate the imaginary term. The rest of the solution was just slog-work.