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Write the given number in polar form $re^{i\theta}$

i) $z = -8\pi (1 + \sqrt{3}i)$

So I thought that $\theta = \arctan(-\sqrt{3}/-1) = \frac{4\pi}{3}$ and it would be $z = 8\pi e^{i\frac{4\pi}{3}}$

But there is a factor of $2$ that I am missing, so the correct answer should be $16\pi e^{-i2\pi/3}$ according to my book. I am not sure where $2\pi/3$ comes from.

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  • $\begingroup$ By convention, $-\pi < \theta \leq \pi$. Hence we will convert $\frac{4\pi}{3}$ to the required range by subtracting $2 \pi$ ("one round") $\endgroup$ – Kelvin Soh Aug 4 '14 at 2:57
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$-2 \pi /3$ and $4 \pi/3$ are the same angle, so that part is fine. The factor of 2 which you're missing is due to the fact that $|1+\sqrt{3} i|$ is $2$, not $1$. So you have to pull out a factor of $2$ to see $\frac{1}{2} + \frac{\sqrt{3}}{2} i = e^{i \pi/3}$. Then use the - to get the other $\pi$ angle.

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  • $\begingroup$ Ohhh I forgot about that! Thanks. $\endgroup$ – Hawk Aug 4 '14 at 3:08

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