Discussion
If \(k = 4n + 3\), where \(n\) is an integer and \(i^{2} = -1\) then \(i^{k}\) is equal to
Answer: E
\(i^{k} = -i\).
Powers of \(i\) repeat every four: \(i^{1} = i,\ i^{2} = -1,\ i^{3} = -i,\ i^{4} = 1\). Since \(i^{4n} = (i^{4})^{n} = 1\) for any integer \(n\):
\[ \begin{aligned} i^{4n+3} &= i^{4n} \cdot i^{3} \\ &= 1 \times (-i) = -i \end{aligned} \]
The other values belong to the other remainders: exponent \(4n\) gives 1, \(4n + 1\) gives \(i\), \(4n + 2\) gives \(-1\). Zero is never a power of \(i\).
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