If i = sqrt{-1}, then [i^{18} + (frac{1}{i})^ | vedclass

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(D) Given $i = \sqrt{-1}$, we know $i^2 = -1$, $i^4 = 1$, and $\frac{1}{i} = -i$.First, simplify $i^{18} = (i^4)^4 \cdot i^2 = (1)^4 \cdot (-1) = -1$.

Vedclass · Accepted Answer

$2 - 2i$

Vedclass · Answer

(D) Given $i = \sqrt{-1}$, we know $i^2 = -1$, $i^4 = 1$, and $\frac{1}{i} = -i$.First, simplify $i^{18} = (i^4)^4 \cdot i^2 = (1)^4 \cdot (-1) = -1$.Next, simplify $(\frac{1}{i})^{25} = (-i)^{25} = (-1)^{25} \cdot i^{25} = -1 \cdot (i^4)^6 \cdot i = -1 \cdot 1 \cdot i = -i$.Substitute these into the expression: $[i^{18} + (\frac{1}{i})^{25}]^3 = [-1 - i]^3$.Factor out $-1$: $[-1(1 + i)]^3 = (-1)^3 (1 + i)^3 = -1(1^3 + i^3 + 3(1)(i)(1 + i))$.Since $i^3 = -i$, we have: $-1(1 - i + 3i + 3i^2) = -1(1 + 2i - 3) = -1(-2 + 2i) = 2 - 2i$.

If $i = \sqrt{-1}$, then $[i^{18} + (\frac{1}{i})^{25}]^3 = $

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