Question
Evaluate the following:

$\frac{1}{i^{58}}$

Answer

We know that, $i^2=-1, i^3=-i, i^4=1$

$\frac{1}{\mathrm{i}^{58}}=\frac{1}{\left(\mathrm{i}^4\right)^{14} \mathrm{i}^2}=\frac{1}{(1)^{14}(-1)}=-1$

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