If text{cis } alpha is the common value of (- | vedclass

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(A) Let $z_1 = (-1)^{1/4}$. We can write $-1 = \cos(\pi + 2k\pi) + i\sin(\pi + 2k\pi) = e^{i(\pi + 2k\pi)}$.Thus,$z_1 = e^{i(\frac{\pi}{4} + \frac{k\p

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$-1$

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(A) Let $z_1 = (-1)^{1/4}$. We can write $-1 = \cos(\pi + 2k\pi) + i\sin(\pi + 2k\pi) = e^{i(\pi + 2k\pi)}$.Thus,$z_1 = e^{i(\frac{\pi}{4} + \frac{k\pi}{2})}$ for $k = 0, 1, 2, 3$.The values are $e^{i\pi/4}, e^{i3\pi/4}, e^{i5\pi/4}, e^{i7\pi/4}$.Let $z_2 = (-i)^{1/2}$. We can write $-i = \cos(\frac{3\pi}{2} + 2n\pi) + i\sin(\frac{3\pi}{2} + 2n\pi) = e^{i(\frac{3\pi}{2} + 2n\pi)}$.Thus,$z_2 = e^{i(\frac{3\pi}{4} + n\pi)}$ for $n = 0, 1$.The values are $e^{i3\pi/4}$ and $e^{i7\pi/4}$.The common values are $e^{i3\pi/4}$ and $e^{i7\pi/4}$.For $e^{i3\pi/4}$,$\alpha = \frac{3\pi}{4}$,so $	an \alpha = 	an(\frac{3\pi}{4}) = -1$.For $e^{i7\pi/4}$,$\alpha = \frac{7\pi}{4}$,so $	an \alpha = 	an(\frac{7\pi}{4}) = -1$.Therefore,$	an \alpha = -1$.

If $\text{cis } \alpha$ is the common value of $(-1)^{1/4}$ and $(-i)^{1/2}$,then $\tan \alpha = $

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