If $S(n)=i^n + i^{-n}$ n is a positive integer, then the total number of distinct values of $S(n)$ are I wrote it in the form
$$S(n)=\frac{i^{2n}+1}{i^n}$$
$$=\frac{(-1)^n+1}{i^n}$$
If n is odd, then the value will be zero.
If n is even then it will become 
$$\frac{2}{-1}$$
So I can find only 2 distinct values of $S(n)$ but the answer says there are 3. What is the third value?
 A: We have that


*

*$i^n=i,-1,-i,1,\dots$

*$i^{-n}=-i,-1,i,1,\dots$
and therefore


*

*$S(1)=i^{1} + i^{-1}=i-i=0$

*$S(2)=i^{2} + i^{-2}=-1-1=-2$

*$S(3)=i^{3} + i^{-3}=-i+i=0$

*$S(4)=i^{4} + i^{-4}=1+1=2$

*$S(5)=i^{5} + i^{-5}=S(1)$

*$\dots$
A: 
If n is even then it will become  $$\frac{2}{-1}$$

False. Actually, you need to, in this case, look at $n$ modulo $4$. This is because $i^{4k} = (-1)^4=1$, while $i^{4k+2} = i^{4k}\cdot i^2 = 1\cdot (-1) = -1$
A: $i=e^{i\pi/2}$
Using How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?
$$S(n)=2\cos\dfrac{n\pi}2$$
Now if $S(n)=S(n+m)$
$$2\cos\dfrac{n\pi}2=2\cos\dfrac{(m+n)\pi}2$$
$$\iff4\sin\dfrac{m\pi}4\sin\dfrac{\pi(m+2n)}4=0$$
As $\sin\dfrac{\pi(m+2n)}4$ is not constant, $$\sin\dfrac{m\pi}4=0$$
$$\implies\dfrac{m\pi}4=r\pi\iff m=4r$$
 where $r$ is an integer
So, the fundamental period of $S(n)$ is $4$
A: You can use the Euler identity:
$$ e^{ix} = \cos(x)+i\sin(x) $$
Where for $ x = \pi/2 $ you obtain that $ i = e^{i\frac{\pi}{2}} $. Now from the initial question you derive that:
$$ S(n) = i^n+i^{-n} = (e^{i\frac{\pi}{2}})^n + (e^{i\frac{\pi}{2}})^{-n} = e^{i\frac{n\pi}{2}} + e^{-i\frac{n\pi}{2}} = $$ $$ = \cos(\frac{n\pi}{2}) + i\sin(\frac{n\pi}{2})+\cos(\frac{n\pi}{2}) - i\sin(\frac{n\pi}{2}) = 2\cos(\frac{n\pi}{2})$$
And for $ n = 0,1,2 $ you get the desired three values $ 2,0,-2. $
