Evaluate:

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

$\Rightarrow i^{4 \times 10+1}+i^{-4 \times 18+1}\left(\right.$ Since $\left.i^{4 n+1}=i\right)$

$\Rightarrow \mathrm{i}^{1}+\mathrm{i}^{1}$

$\Rightarrow 2 \mathrm{i}$

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

(ii) $\left(i^{53}+\frac{1}{i^{53}}\right)$

$\Rightarrow i^{53}+i^{-53}$

$\Rightarrow i^{4 \times 13+1}+i^{-4 \times 14+3}$ (Since $i^{4 n+1}=i$

$\left.\Rightarrow i^{1}+i^{3} i^{4 n+3}=-1\right)$

$\Rightarrow 0$

Hence, $\left(i^{53}+\frac{1}{i^{53}}\right)=0$