If z = sum_{n=0}^{2026} i^n, where i = sqrt{- | vedclass

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(A) The sum is a geometric series: $z = \sum_{n=0}^{2026} i^n = \frac{1(1 - i^{2027})}{1 - i}$.Since $i^4 = 1$, we have $i^{2027} = i^{2024} \cdot i^3

Vedclass · Accepted Answer

$e^{i\frac{\pi}{4}}$

Vedclass · Answer

(A) The sum is a geometric series: $z = \sum_{n=0}^{2026} i^n = \frac{1(1 - i^{2027})}{1 - i}$.Since $i^4 = 1$, we have $i^{2027} = i^{2024} \cdot i^3 = 1 \cdot (-i) = -i$.Thus, $z = \frac{1 - (-i)}{1 - i} = \frac{1 + i}{1 - i}$.Multiply numerator and denominator by $(1 + i)$: $z = \frac{(1 + i)^2}{(1 - i)(1 + i)} = \frac{1 + 2i + i^2}{1 - i^2} = \frac{1 + 2i - 1}{1 + 1} = \frac{2i}{2} = i$.We need to find $\sqrt{z} = \sqrt{i}$.Since $i = e^{i\frac{\pi}{2}}$, then $\sqrt{i} = (e^{i\frac{\pi}{2}})^{1/2} = e^{i\frac{\pi}{4}}$.Therefore, one value is $e^{i\frac{\pi}{4}}$.

If $z = \sum_{n=0}^{2026} i^n$, where $i = \sqrt{-1}$, then one of the values of $\sqrt{z}$ is...

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