Prove that

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.