Question:
Evaluate: $\left(i^{41}+\frac{1}{i^{71}}\right)$
Solution:
We have $\left(i^{41}+\frac{1}{i^{71}}\right)$
$\mathrm{i}^{41}=\mathrm{i}^{40} \cdot \mathrm{i}=\mathrm{i}$
$\mathrm{i}^{71}=\mathrm{i}^{68} \cdot \mathrm{i}^{3}=-\mathrm{i}$
Therefore
$\left(i^{41}+\frac{1}{i^{71}}\right)=i-\frac{1}{i}=\frac{i^{2}-1}{i}$
$\left(i^{41}+\frac{1}{i^{71}}\right)=-\frac{2}{i} \times \frac{i}{i}$
$\left(i^{41}+\frac{1}{i^{71}}\right)=-\frac{2 i}{i^{2}}=2 i$
Hence, $\left(i^{41}+\frac{1}{i^{71}}\right)=2 i$
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