Tính
a) \({{{\left( {1 + i} \right)}^{2006}}}\)
b) \({{{\left( {1 - i} \right)}^{2006}}}\)
a)
\(\begin{array}{l}
{(1 + i)^{2006}} = {[{(1 + i)^2}]^{1003}} = {(1 + 2i - 1)^{1003}}\\
= {(2i)^{1003}} = {2^{1003}}.{({i^2})^{501}}.i = - {2^{1003}}i
\end{array}\)
b)
\(\begin{array}{l}
{(1 - i)^{2006}} = {[{(1 - i)^2}]^{1003}} = {(1 - 2i - 1)^{1003}}\\
= {( - 2i)^{1003}} = - {2^{1003}}.{({i^2})^{501}}.i = {2^{1003}}i
\end{array}\)
-- Mod Toán 12
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