If z = frac{sqrt{3}}{2} + frac{i}{2} where i | vedclass

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(A) Given $z = \frac{\sqrt{3}}{2} + \frac{i}{2} = \cos(\frac{\pi}{6}) + i \sin(\frac{\pi}{6}) = e^{i\pi/6}$.We need to evaluate $(1 + iz + z^5 + iz^8)

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(A) Given $z = \frac{\sqrt{3}}{2} + \frac{i}{2} = \cos(\frac{\pi}{6}) + i \sin(\frac{\pi}{6}) = e^{i\pi/6}$.We need to evaluate $(1 + iz + z^5 + iz^8)^9$.First,simplify the expression inside the parenthesis:$1 + iz + z^5 + iz^8 = 1 + i(e^{i\pi/6}) + (e^{i\pi/6})^5 + i(e^{i\pi/6})^8$$= 1 + e^{i\pi/2}e^{i\pi/6} + e^{i5\pi/6} + e^{i\pi/2}e^{i8\pi/6}$$= 1 + e^{i2\pi/3} + e^{i5\pi/6} + e^{i11\pi/6}$$= 1 + (-\frac{1}{2} + i\frac{\sqrt{3}}{2}) + (-\frac{\sqrt{3}}{2} + i\frac{1}{2}) + (\frac{\sqrt{3}}{2} - i\frac{1}{2})$$= 1 - \frac{1}{2} + i\frac{\sqrt{3}}{2} = \frac{1}{2} + i\frac{\sqrt{3}}{2} = e^{i\pi/3}$.Now,raise this to the power of $9$:$(e^{i\pi/3})^9 = e^{i3\pi} = \cos(3\pi) + i \sin(3\pi) = -1 + 0 = -1$.

If $z = \frac{\sqrt{3}}{2} + \frac{i}{2}$ where $i = \sqrt{-1}$,then $(1 + iz + z^5 + iz^8)^9$ is equal to:

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