Discussion
If \(\alpha\) is a real number satisfying \(\alpha^{2} - \dfrac{1}{\alpha^{2}} = 2\), then the value of \(\left(\alpha + \dfrac{i}{\alpha}\right)^{16}\) is equal to
Answer: B
The value is 4096.
Square the bracket first:
\[ \begin{aligned} \left(\alpha + \frac{i}{\alpha}\right)^{2} &= \alpha^{2} + 2i + \frac{i^{2}}{\alpha^{2}} \\ &= \left(\alpha^{2} - \frac{1}{\alpha^{2}}\right) + 2i = 2 + 2i \end{aligned} \]
\[ \begin{aligned} (2 + 2i)^{2} &= 4 + 8i + 4i^{2} = 8i \\ \left(\alpha + \tfrac{i}{\alpha}\right)^{16} &= (2 + 2i)^{8} = (8i)^{4} \\ &= 4096\,i^{4} = 4096 \end{aligned} \]
The option 2048 is half the answer, and 4048 and 5096 are look-alike numbers; \(8^{4} = 4096\) is the only power that fits.
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