Prove that $\frac{1}{i}-\frac{1}{i^{2}}+\frac{1}{i^{3}}-\frac{1}{i^{4}}=0$
Given: $\frac{1}{\mathrm{i}}-\frac{1}{\mathrm{i}^{2}}+\frac{1}{\mathrm{i}^{3}}-\frac{1}{\mathrm{i}^{4}}$
To prove : $\frac{1}{\mathrm{i}}-\frac{1}{\mathrm{i}^{2}}+\frac{1}{\mathrm{i}^{3}}-\frac{1}{\mathrm{i}^{4}}=0$.
$\Rightarrow$ L.H.S. $=\mathrm{i}^{-1}-\mathrm{i}^{-2}+\mathrm{i}^{-3}-\mathrm{i}^{-4}$
$\Rightarrow \mathrm{i}^{-4 \times 1+3}-\mathrm{i}^{-4 \times 1+2}+\mathrm{i}^{-4 \times 1+3}-\mathrm{i}^{-4 \times 1}$
Since $i^{4 n}=1$
$\Rightarrow i^{4 n+1}=i$
$\Rightarrow \mathrm{i}^{4 \mathrm{n}+2}=-1$
$\Rightarrow i^{4 n+3}=-1$
So,
$\Rightarrow \mathrm{i}^{1}-\mathrm{i}^{2}+\mathrm{i}^{3}-1$
$\Rightarrow \mathrm{i}+1-\mathrm{i}-1$
$\Rightarrow 0$
$\Rightarrow$ L.H.S $=$ R.H.S
Hence Proved
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.