Evaluate:

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, $\left(i^{41}+\frac{1}{i^{71}}\right)=2 i$