Question:
Prove that $\left\{\mathrm{i}^{21}-\left(\frac{1}{\mathrm{i}}\right)^{46}\right\}^{2}=2 \mathrm{i}$.
Solution:
L.H.S. $=\left\{i^{21}-\left(\frac{1}{i}\right)^{46}\right\}^{2}$
$=\left\{i^{4 \times 5+1}-i^{-4 \times 12+2}\right\}^{2}$
Since $i^{4 n}=1$
$i^{4 n+1}=i$
$i^{4 n+2}=i^{2}=-1$
$i^{4 n+3}=i^{3}=-1$
$=\left\{i^{1}-i^{2}\right\}^{2}$
$=\{i+1\}^{2}$
Now, applying the formula $(a+b)^{2}=a^{2}+b^{2}+2 a b$
$=i^{2}+1+2 i .$
$=-1+1+2 i$
$=2 i$
L.H.S = R.H.S
Hence proved.