What Does (1+√3i)^n + (1-√3i)^n Equal for Certain n?

[zorro]

Homework Statement

If n is an integer which leaves remainder one when divided by three, then (1+√3i)ⁿ + (1-√3i)ⁿ equals

a) -2ⁿ⁺¹
b) 2ⁿ⁺¹
c) -(-2)ⁿ
d) -2ⁿ

The Attempt at a Solution

Multiply and divide each of the two expression inside bracket by 2.
2ⁿ(cosπ/3 + isinπ/3)ⁿ + 2ⁿ(cosπ/3-isinπ/3)ⁿ
On simplifiying, the expression equals
2ⁿ(2cosnπ/3)=2ⁿ(-1)=-2ⁿ which is same as (d)

The correct option is (c). I have no idea where I went wrong.

 

[Dick]

Put n=1 into 2^n(2cosnπ/3). cos(pi/3) isn't equal to (-1/2). It's +1/2. Can you take it from there?

 

[zorro]

n=1 does not give 1 as a remainder when divided by 3 does it?
I put n=4 in the above expression. The answer from your idea is not even there in the options.

 

[eumyang]

n=1 does not give 1 as a remainder when divided by 3 does it?

Uh, yes it does. 1 divided by 3 is 0 with a remainder of 1.

 

[Dick]

n=1 does not give 1 as a remainder when divided by 3 does it?
I put n=4 in the above expression. The answer from your idea is not even there in the options.

What idea? I'm just trying to get you to realize 2*cos(n*pi/3) isn't always equal to (-1) when n=1,4,7,10,...

 

[zorro]

What idea? I'm just trying to get you to realize 2*cos(n*pi/3) isn't always equal to (-1) when n=1,4,7,10,...

Sorry I did not get your point earlier. I realize that for n=1,7,13... cos(npi/3) is +1/2. How do I use this point to get -(-2)ⁿ ?

 

[Dick]

Sorry I did not get your point earlier. I realize that for n=1,7,13... cos(npi/3) is +1/2. How do I use this point to get -(-2ⁿ) ?

Ok, and for n=4,10,16,... it's -1/2. Can't you write that result as c*(-1)^n for some constant c?

 

[zorro]

2ⁿ(2cosnπ/3) = 2ⁿ[2*1/2*(-1)ⁿ⁺¹] = -(-2)ⁿ
I wonder if there is some other easier method.

 

[SammyS]

Let n = 3k+1, where k is an integer.

$$\left(1+i\sqrt{3}\right)^{3}=\left(1-i\sqrt{3}\right)^{3}=-8=(-2)^3$$

 

[zorro]

I figured out a much easier way.
(1+√3i)ⁿ + (1-√3i)ⁿ = 2ⁿ(-w²)ⁿ + 2ⁿ(-w)ⁿ = (-2ⁿ)(w²+w) = -(-2)ⁿ, where w is the complex cube root of unity.

 

[SammyS]

Let n = 3k+1, where k is an integer.

$$\left(1+i\sqrt{3}\right)^{3}=\left(1-i\sqrt{3}\right)^{3}=-8=(-2)^3$$

This is what I had in mind for the above:

n = 3k+1 → 3k = n-1

$$\left(1+i\sqrt{3}\right)^{3k+1}=\left(1+i\sqrt{3}\right)\left(\left(1+i\sqrt{3}\right)^{3}\right)^k$$

$$=\left(1+i\sqrt{3}\right)\left(-2\right)^{3k}$$

$$=\left(1+i\sqrt{3}\right)\left(-2\right)^{n-1}$$​

Similarly:
$$\left(1-i\sqrt{3}\right)^{3k+1}=\left(1-i\sqrt{3}\right)\left(-2\right)^{n-1}$$

Adding these two results gives: -(-2)⁽ⁿ⁾

 

[zorro]

Nice one!