The product of the four values of the complex | vedclass

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(B) Let $z = (1+i)^{3/4}$. First,express $1+i$ in polar form: $1+i = \sqrt{2} e^{i(\pi/4 + 2k\pi)}$.Then,$z = (\sqrt{2})^{3/4} e^{i \frac{3}{4}(\pi/4

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$2(1-i)$

Vedclass · Answer

(B) Let $z = (1+i)^{3/4}$. First,express $1+i$ in polar form: $1+i = \sqrt{2} e^{i(\pi/4 + 2k\pi)}$.Then,$z = (\sqrt{2})^{3/4} e^{i \frac{3}{4}(\pi/4 + 2k\pi)} = 2^{3/8} e^{i(\frac{3\pi}{16} + \frac{3k\pi}{2})}$ for $k = 0, 1, 2, 3$.The product of these four values is $P = \prod_{k=0}^{3} 2^{3/8} e^{i(\frac{3\pi}{16} + \frac{3k\pi}{2})} = (2^{3/8})^4 e^{i \sum_{k=0}^{3} (\frac{3\pi}{16} + \frac{3k\pi}{2})}$.$P = 2^{3/2} e^{i (4 \cdot \frac{3\pi}{16} + \frac{3\pi}{2} \cdot \frac{3(4)}{2})} = 2^{3/2} e^{i (\frac{3\pi}{4} + 9\pi)} = 2^{3/2} e^{i (\frac{3\pi}{4} + \pi)} = 2^{3/2} e^{i (7\pi/4)}$.Since $e^{i(7\pi/4)} = \cos(7\pi/4) + i \sin(7\pi/4) = \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} = \frac{1-i}{\sqrt{2}}$.$P = 2^{3/2} \cdot \frac{1-i}{\sqrt{2}} = 2^1 (1-i) = 2(1-i)$.

The product of the four values of the complex number $(1+i)^{3/4}$ is

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