# Complex numbers expression

## 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
Homework Helper
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?

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
Homework Helper
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
Homework Helper
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,...

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 realise that for n=1,7,13.... cos(npi/3) is +1/2. How do I use this point to get -(-2)n ?

Last edited:
Dick
Homework Helper
Sorry I did not get your point earlier. I realise 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?

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

SammyS
Staff Emeritus
Homework Helper
Gold Member
Let n = 3k+1, where k is an integer.

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

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
Staff Emeritus
Homework Helper
Gold Member
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)

Nice one!