# Complex number - arctan

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

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!

Since $$\arctan\left(-\sqrt3\right)=-\frac\pi3$$, $$\arctan\left(-\sqrt3\right)+\pi=\frac{2\pi}3$$.
• 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. Jun 1, 2020 at 8:57
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 $$\tan^{-1}\sqrt3=\frac{\pi}{3}$$
So, if $$\tan(x)=y$$, it follows that $$\arctan(y):=\tan^{-1}(y)=\tan^{-1}(\tan(x))=x$$
In this case, we have $$\arctan(-\sqrt3)=\tan^{-1}(-\sqrt3)=-\pi/3$$