Cartesian to polar confusion (simple)?

[noahsdev]

Homework Statement

Convert -2+2√3i to polar coordinates.

Homework Equations

r = √x²+y²
θ = tan^-1(y/x)

The Attempt at a Solution

I am confused because θ = tan^-1(2√3/2) = tan^-1(√3) = -π/3 and r = 4, so that would make the polar form 4cis(-π/3), but the calculator gives: 4cis(2π/3).
I think the calculator is right because when I convert my answer (4cis(-π/3)) back to cartesian it gives -2-2√3i, whereas the other (4cis(2π/3))gives the right answer, -2+2√3i.

Can someone explain what I'm doing wrong?
Thanks. :)

 

[arildno]

In which quadrant does your complex number lie?
For which interval of angles is the standard tangent function defined?

 

[HallsofIvy]

Tangent is, of course, periodic and your calculator can give only one value- the "principal" value which, for tangent, is the value of $\theta$ with the smallest absolute value. Since tangent is periodic with period $\pi$, $tan(-\pi/3)= tan(-\pi/3+ \pi)= tan(2\pi/3)$.

You distinguish between them by noting that $-\pi/3$ is in the fourth quadrant, (+,-), while $2\pi/3$ is in the second quadrant, (-, +).

 

[noahsdev]

In which quadrant does your complex number lie?
For which interval of angles is the standard tangent function defined?

OK, I have found the angle using x and y (cos and sin) and they both confirm that the calculator is correct. And yes, it does make sense since the complex number lies in quadrant 1 but why is the tan function wrong? I'm guessing you were hinting at that part but I really don't know. :)

 

[SteamKing]

OK, I have found the angle using x and y (cos and sin) and they both confirm that the calculator is correct. And yes, it does make sense since the complex number lies in quadrant 1 but why is the tan function wrong? I'm guessing you were hinting at that part but I really don't know. :)

Are you sure that -2+2SQRT(3)i is in the first quadrant? Why don't you make a sketch?

 

[noahsdev]

Tangent is, of course, periodic and your calculator can give only one value- the "principal" value which, for tangent, is the value of $\theta$ with the smallest absolute value. Since tangent is periodic with period $\pi$, $tan(-\pi/3)= tan(-\pi/3+ \pi)= tan(2\pi/3)$.

You distinguish between them by noting that $-\pi/3$ is in the fourth quadrant, (+,-), while $2\pi/3$ is in the second quadrant, (-, +).

Yes that makes sense. Thanks.
P.S I know the quadrants haha I misstyped :)