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1957 AHSME Problems/Problem 42

Problem 42

If $S = i^n + i^{-n}$ , where $i = \sqrt{-1}$ and $n$ is an integer, then the total number of possible distinct values for $S$ is:

$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 3\qquad \textbf{(D)}\ 4\qquad \textbf{(E)}\ \text{more than 4}$

Solution

We first use the fact that $i^{-n}=\frac1{i^n}=\left(\frac1i\right)^n=(-i)^n$ . Note that $i^4=1$ and $(-i)^4=1$ , so $i^n$ and $(-i)^n$ have are periodic with periods at most 4. Therefore, it suffices to check for $n=0,1,2,3$ .

For $n=0$ , we have $i^0+(-i)^0=1+1=2$ .

For $n=1$ , we have $i^1+(-i)^1=i-i=0$ .

For $n=2$ , we have $i^2+(-i)^2=-1-1=-2$ .

For $n=3$ , we have $i^3+(-i)^3=-i+i=0$ .

Hence, the answer is $\boxed{\textbf{(C)}\ 3}$ .

Solution 2

Notice that the powers of $i$ cycle in cycles of 4. So let's see if $S$ is periodic.

For $n=0$ : we have $2$ .

For $n=1$ : we have $0$ .

For $n=2$ : we have $-2$ .

For $n=3$ : we have $0$ .

For $n=4$ : we have $2$ again. Well, it can be seen that $S$ cycles in periods of 4. Select $\boxed{C}$ .

~hastapasta

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