Advanced High-School Mathematics

100 CHAPTER 2 Discrete Mathematics

That is to say, the real and imaginary parts of (a+bi)nare cosnθ

and sinnθ, whereθ is as above. From this, one finally concludes

that the solution of (2.4) in the present case has the form

un = Acosnθ+Bsinnθ, n= 0, 1 , 2 ,...,

whereAandBare real constants.

It’s time to come up for air and look at an example.

Example 2. Solve the second-order homogeneous difference equa-

tion

un+2=−un+1−un, n= 0, 1 , 2 ,..., (2.6)

whereu 0 = 1, u 1 = 1.

Solution. The characteristic polynomialC(x) =x^2 +x+1 which

has zeros

−1 +i

√

3

2

and

− 1 −i

√

3

2

. We write the first complex
number in trigonometric form

−1 +i

√

3

2

= cos

2 π

3

+isin

2 π

3

,

from which it follows that

Ñ

−1 +i

√

3

2

én

= cos

2 πn

3

+isin

2 πn

3

.

From this it follows that the general solution is given by

un = Acos

2 πn

3

+Bsin

2 πn

3

, n= 0, 1 , 2 ,....