prove Demoivre's Theorem

Question

prove Demoivre's Theorem using induction

Answer 1

Prove that

(cos(x) + i sin(x))ⁿ = cos(nx) + i sin(nx),

 

where i =√-1.

 

The inductive proof has two steps.

 

Step 1.

n=1:

 

cos(x) + i sin(x) = (cos(x) + i sin(x))¹ = cos(1x) + i sin(1x) = cos(x) + i sin(x)

So the general statement for the proposition is true for n = 1.

 

Step 2.

Let the proposition is true for some natural number k>=1:

(cos(x) + i sin(x))^k= cos(kx) + i sin(kx)

 

We will prove that

(cos(x) + i sin(x))^k+1 = cos((k+1)x) + i sin((k+1)x.

 

Proof:

(cos(x) + i sin(x))^k+1 =(cos(x) +i sin(x))^k(cos(x) + i sin(x))                                                          (1)

                                =(cos(kx) +i sin(kx))(cos(x) + i sin(x))                                                         (2)

                                = cos(kx) cos(x) + i(sin(kx)cos(x)) +i(cos(kx)sin(x)) + i²(sin(kx)sin(x))          (3)

                                = cos(kx) cos(x) - sin(kx)sin(x) + i[sin(kx)cos(x) +sin(x)cos(kx)]                   (4)

                                =cos((kx +x) + i sin(kx+x)                                                                           (5)

                                =cos((k+1)x) + i sin((k+1)x)                                                                        (6)

 

We used these facts:

1. when we multiply binomial by binomial we multiply each of the addends in the first parenthesis by each term in second parenthesis (2),
2. we combined like terms (3)
3. We used the fact that i²=-1, (4)
4. We used the formula for cosine of sum of angles and sine of sum of angles (5)
5. We simplified (kx+x) to look as in the proposition.(6)