1
$\begingroup$

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

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!

enter image description here

$\endgroup$
1

3 Answers 3

Reset to default
2
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – user12595983
    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. $\endgroup$
    – José Carlos Santos
    Jun 1, 2020 at 8:57
1
$\begingroup$

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 $\tan^{-1}\sqrt3=\frac{\pi}{3}$

$\endgroup$
1
$\begingroup$

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 $\arctan(-\sqrt3)=\tan^{-1}(-\sqrt3)=-\pi/3$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.