Show that:∀n∈N([(2+i)n+(2−i)n]∈R)

Question

Zernerqcw · Accepted Answer

There are two ways to write a complex number: rectangular form, e.g.,

x + i y

, and polar form, e.g.,

r e^{i θ}

. The conversion between them uses trig functions:

r e^{i θ} = r \cos θ + i r \sin θ

(1)
Going in the other direction,

x + i y = \sqrt{x^{2} + y^{2}} e^{i θ}

where

θ

is any angle such that

\cos θ = \frac{x}{\sqrt{x^{2} + y^{2}}}

and

\sin θ = \frac{y}{\sqrt{x^{2} + y^{2}}}

The important thing for your argument is that

r = \sqrt{x^{2} + y^{2}}

The r corresponding to

2 + i

is therefore

\sqrt{2^{2} + 1^{2}} = \sqrt{5}

, and that corresponding to

2 - i

is

\sqrt{2^{2} + {(- 1)}^{2}} = \sqrt{5}

as well. The angles for

2 + i

is an angle

θ

whose cosine is

\frac{2}{\sqrt{5}}

and whose sine is

\frac{1}{\sqrt{5}}

, while the angle for

2 - i

is an angle whose cosine is

\frac{2}{\sqrt{5}}

and whose sine is

- \frac{1}{\sqrt{5}}

. It doesn’t matter exactly what they are; the important thing is that if we let the first be

θ

, the second is

- θ

, since

\cos (- θ) = \cos θ

and

\sin (- θ) = - \sin θ

Substituting into (1) gives you

2 + i = \sqrt{5} \cos θ + i \sqrt{5} \sin θ = \sqrt{5} (\cos θ + i \sin θ) = \sqrt{5} e^{i θ}

and

2 - i = \sqrt{5} \cos (- θ) + i \sqrt{5} \sin (- θ) = \sqrt{5} (\cos θ - i \sin θ) = \sqrt{5} e^{- i θ}

Now use the fact that it’s easy to raise an exponential to a power:

{(2 + i)}^{n} + {(2 - i)}^{n} = {(\sqrt{5})}^{n} {(e^{i θ})}^{n} + {(\sqrt{5})}^{n} {(e^{- i θ})}^{n}

= {(\sqrt{5})}^{n} (e^{\in θ} + e^{- i n θ})

This is helpful 0 Flag

Show that: \forall n\in\mathbb{N}([(2+i)^n+(2-i)^n]\in\mathbb{R})