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$$z=-2+2\sqrt{3}i\implies x=-2, y=2\sqrt3$$ $$r=\sqrt{x^2+y^2}=\sqrt{4+12}=4$$ $$\text{Angle}=\arctan\left(\frac{2\sqrt3}{-2}\right)+\pi=\frac{2\pi}{3}=120^\circ$$

1) May I know how $\arctan\left(\dfrac{2\sqrt3}{-2}\right)+\pi$ turns into $\dfrac{2\pi}{3}$?

2) Can I use calculator to do the calculation and how?

Thanks for the kindness!

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3 Answers 3

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Since $\arctan\left(-\sqrt3\right)=-\frac\pi3$, $\arctan\left(-\sqrt3\right)+\pi=\frac{2\pi}3$.

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  • $\begingroup$ i didn't know arctan(-{\sqrt 3})= \frac{−π}{3}, may i know what keywords should i google to learn? Thanks for your helpful answer. $\endgroup$ Jun 1, 2020 at 8:50
  • $\begingroup$ WHat I wrote was that $\arctan\left(-\sqrt3\right)=-\frac\pi3$. It follows from the fact that $\arctan$ is an odd function and that $\tan\left(\frac\pi3\right)=\sqrt3$. You can learn much more at the Wikipedia article on inverse trigonometric functions. $\endgroup$ Jun 1, 2020 at 8:57
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Notice, $\tan(-\theta)=-\tan\theta$ $$\therefore \arctan\left(\frac{2\sqrt3}{-2}\right)+\pi=\arctan\left(-\sqrt{3}\right)+\pi=-\frac{\pi}{3}+\pi=\frac{2\pi}{3}=120^\circ$$ You can use calculator to find amplitude & argument. Do remember $\tan^{-1}\sqrt3=\frac{\pi}{3}$

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First of all, you need to know that arctan is the inverse function of tangent.

So, if $\tan(x)=y$, it follows that $\arctan(y):=\tan^{-1}(y)=\tan^{-1}(\tan(x))=x$

In this case, we have $\arctan(-\sqrt3)=\tan^{-1}(-\sqrt3)=-\pi/3$

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