The value of {i^{22}-(frac{1}{i})^{35}}^2 is | vedclass

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(A) We know that $i^2 = -1$,$i^4 = 1$. First,calculate $i^{22}$: $i^{22} = (i^4)^5 	imes i^2 = 1^5 	imes (-1) = -1$. Next,calculate $(\frac{1}{i})^{

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$2i$

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(A) We know that $i^2 = -1$,$i^4 = 1$. First,calculate $i^{22}$: $i^{22} = (i^4)^5 	imes i^2 = 1^5 	imes (-1) = -1$. Next,calculate $(\frac{1}{i})^{35}$: $\frac{1}{i} = -i$. So,$(-i)^{35} = -(i^{35}) = -(i^{32} 	imes i^3) = -(1 	imes -i) = i$. Now,substitute these values into the expression: $\{i^{22} - (\frac{1}{i})^{35}\}^2 = \{-1 - i\}^2$. Expand the square: $(-1 - i)^2 = (-1)^2 + (-i)^2 + 2(-1)(-i) = 1 + i^2 + 2i$. Since $i^2 = -1$,we get $1 - 1 + 2i = 2i$.

The value of $\{i^{22}-(\frac{1}{i})^{35}\}^2$ is

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