Question

(a) Prove that
(−1 + i
√
3)n + (−1 − i
√
3)n
has either the value 2n+1 or the value −2
n
if n is any integer (positive, negative or zero).

Answer 1

(-1+i√3)$^n$ + (-1-i√3)$^n$

= 2$^n$ ($\frac{-1+i√3}{2})^n$ + 2$^n$ ($\frac{-1-i√3}{2})^n$

= $2^n$ $(\frac{-1}{2}+\frac{i√3}{2})^n$ + $2^n$ $(\frac{-1}{2}-\frac{i√3}{2})^n$

= $2^n$ ( cos $\frac{2π}{3}+i$ sin$\frac{2π}{3}$ )$^n$ + 2$^n$ ( cos$\frac{2π}{3}-i$sin$\frac{2π}{3}$ )$^n$

= $2^n$ ( cos $\frac{2nπ}{3}+$ isin$\frac{2nπ}{3}$ ) + $2^n$ ( cos$\frac{2nπ}{3}-$ isin$\frac{2nπ}{3}$ ) by D'Moivres theorem

= $2^n$ . 2 cos$\frac{2nπ}{3}$

= $2^{n+1}$ cos$\frac{2nπ}{3}$

Now as n is an integer, n can be any one of the forms 3k, 3k+1, 3k+2 where k is an integer.

When n = 3k

$cos \frac{2nπ}{3}$ = $cos \frac{2(3k)π}{3}$ = cos 2kπ = 1

So $(-1+i√3)^n+(-1-i√3)^n$

= 2$^{n+1}$

When n = 3k+1

$cos \frac{2nπ}{3}$ = $cos \frac{2(3k+1)π}{3}$ = cos(2kπ+$\frac{2π}{3})$ = cos$\frac{2π}{3} = cos(π-\frac{π}{3})= -cos\frac{π}{3}$ = $-\frac{1}{2}$

So $(-1+i√3)^n+(-1-i√3)^n$

= 2$^{n+1}(-\frac{1}{2})$

= $-2^n$

When n = 3k+2

$cos \frac{2nπ}{3}$ = $cos \frac{2(3k+2)π}{3}$ = cos(2kπ +$\frac{4π}{3})$ = cos$\frac{4π}{3}$ = cos(π+$\frac{π}{3}$ )= -cos$\frac{π}{3}$ = $-\frac{1}{2}$

So $(-1+i√3)^n+(-1-i√3)^n$

= $2^{n+1}(-\frac{1}{2})$

= $-2^n$

Finally we can conclude that

$(-1+i√3)^n+(-1-i√3)^n$

= 2$^{n+1}$ or $-$ 2$^{n}$ if n is an integer (positive, negative or zero)