Discussion
The value of \((1 + i)^{10}\) is equal to
Answer: D
\((1 + i)^{10} =\) \(32i\).
Square first, since \((1 + i)^{2}\) is simple:
\[ \begin{aligned} (1 + i)^{2} &= 1 + 2i + i^{2} = 2i \\ (1 + i)^{10} &= (2i)^{5} = 32\,i^{5} = 32i \end{aligned} \]
because \(i^{5} = i^{4}\cdot i = i\). The modulus check agrees: \(|1 + i|^{10} = (\sqrt{2})^{10} = 32\). The options 16 and \(16i\) have the wrong modulus, and 32 or 64 lose the factor \(i\).
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