The amplitude (argument) of (1+i)^{5} is | vedclass

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(B) Let $z = (1+i)^{5}$.First,express $1+i$ in polar form: $1+i = \sqrt{2}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})$.Using De Moivre's theorem,$z =

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$\frac{-3 \pi}{4}$

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(B) Let $z = (1+i)^{5}$.First,express $1+i$ in polar form: $1+i = \sqrt{2}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})$.Using De Moivre's theorem,$z = (\sqrt{2})^{5}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})^{5} = 4\sqrt{2}(\cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4})$.The argument of $z$ is $\frac{5\pi}{4}$.Since the principal argument must lie in the interval $(-\pi, \pi]$,we subtract $2\pi$ from $\frac{5\pi}{4}$:$\frac{5\pi}{4} - 2\pi = -\frac{3\pi}{4}$.Thus,the principal amplitude is $-\frac{3\pi}{4}$.

The amplitude (argument) of $(1+i)^{5}$ is

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