If (1 + isqrt{3})^9 = a + ib,then b is equal | vedclass

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(C) We express the complex number in polar form: $1 + i\sqrt{3} = 2\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = 2\left(\cos\frac{\pi}{3} + i\sin\f

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(C) We express the complex number in polar form: $1 + i\sqrt{3} = 2\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right) = 2e^{i\pi/3}$.Using De Moivre's theorem: $(1 + i\sqrt{3})^9 = (2e^{i\pi/3})^9 = 2^9 \cdot e^{i(3\pi)}$.Since $e^{i(3\pi)} = \cos(3\pi) + i\sin(3\pi) = -1 + i(0) = -1$,we have $(1 + i\sqrt{3})^9 = 2^9(-1) = -512$.Comparing this with $a + ib$,we get $a = -512$ and $b = 0$.

If $(1 + i\sqrt{3})^9 = a + ib$,then $b$ is equal to

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