The number of real values of (-1-sqrt{3} i)^{ | vedclass

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(C) Let $z = -1 - \sqrt{3}i$. First,express $z$ in polar form: $|z| = \sqrt{(-1)^2 + (-\sqrt{3})^2} = \sqrt{1+3} = 2$. $	heta = 	ext{arg}(z) = \pi +

Vedclass · Accepted Answer

$2$

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(C) Let $z = -1 - \sqrt{3}i$. First,express $z$ in polar form: $|z| = \sqrt{(-1)^2 + (-\sqrt{3})^2} = \sqrt{1+3} = 2$. $	heta = 	ext{arg}(z) = \pi + 	an^{-1}(\frac{\sqrt{3}}{1}) = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$. So,$z = 2e^{i(4\pi/3 + 2k\pi)}$ for $k = 0, 1, 2, 3$. Then $z^{3/4} = (2e^{i(4\pi/3 + 2k\pi)})^{3/4} = 2^{3/4} e^{i(3/4)(4\pi/3 + 2k\pi)} = 2^{3/4} e^{i(\pi + 3k\pi/2)}$. For the value to be real,the imaginary part must be zero,i.e.,$\sin(\pi + \frac{3k\pi}{2}) = 0$. This implies $\pi + \frac{3k\pi}{2} = n\pi$ for some integer $n$. Dividing by $\pi$,we get $1 + \frac{3k}{2} = n$,or $2 + 3k = 2n$. Since $2n$ is even and $2$ is even,$3k$ must be even,which means $k$ must be even. For $k \in \{0, 1, 2, 3\}$,the even values are $k = 0$ and $k = 2$. For $k=0$,$	ext{arg} = \pi$,$e^{i\pi} = -1$ (real). For $k=2$,$	ext{arg} = \pi + 3\pi = 4\pi$,$e^{i4\pi} = 1$ (real). Thus,there are $2$ real values.

The number of real values of $(-1-\sqrt{3} i)^{3/4}$ is

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