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[agata510]

\(\displaystyle{ (1+i)^{-6}}\)

[pawellogrd]

\(\displaystyle{ (1+i)^{-6} = \frac{1}{(1+i)^6} = \frac{1}{\left( \left( 1+i\right) ^2\right)^3} = \frac{1}{(1+2i+i^2)^3} = \frac{1}{(1+2i-1)^3} = \frac{1}{(2i)^3} = \frac{1}{8i^3} = \frac{1}{8i \cdot i^2} = \frac{1}{-8i} = -\frac{1}{8i} = -\frac{i}{8i^2} = - \frac{i}{-8} = \frac{i}{8}}\)