# How can $r$ be negative when dealing with polar coordinates?

If by definition $r=\sqrt{x^2 + y^2}$, then why do we allow $r$ to be negative? Relatedly, I do not understand the last section of this conversation discussing points being represented by multiple $\theta$:

Student: So a single point could have many different values?

Mentor: Correct! The values for $r$ can be given as positive and negative values and $\theta$ can be given not only in positive and negative values, but also as any value $\theta$ + any multiple of $2\pi$. So, unlike the Cartesian system where each point has a unique set of coordinates, in the polar system any point can have an infinite number of coordinates!

Student: That means that the point given as $(2,\pi/4)$ could also be given as $(2,-7\pi/4)$ or $(2, 9\pi/4)$ or $(-2,5\pi/4)$!

Mentor: Exactly. Try plotting those points using the Polar Coordinates activity and verify they are the same point!

This part in particular is confusing to me:

$\theta$ can be given not only in positive and negative values, but also as any value $\theta$ + any multiple of 2π.

So if $\theta=5\pi/3$ then it would also equal $5\pi/3 + \pi$? I'm pretty confused on polar coordinates; they seem so inferior to traditional Cartesian when not dealing with imaginary numbers. Could someone clarify this redundancy in polar coordinates?

• Polar coordinates in imaginary numbers are similer to those of trigonometry functions. As $\sin(x) = \sin(x+2k\pi)$. Which means that a point can have infinte number of values. I dont exactly understand your confusion. – Eminem Aug 9 '15 at 9:53

By definition, a point with polar coordinates $r,\theta$ has position: $$x = r\cos \theta$$ $$y = r\sin \theta$$
Inverting this equation gives you many solutions, as you mentioned. Specifically for $r$, you can square both equations and add them up to get: $$x^2 = r^2 \cos^2 \theta$$ $$y^2 = r^2 \sin^2 \theta$$ $$x^2+y^2 = r^2(\cos^2 \theta + \sin^2 \theta) = r^2$$ From this, you can determine that $r$ can be either of the following: $$\sqrt{x^2+y^2},\quad -\sqrt{x^2+y^2}$$ Which one is true? If we know that $\cos \theta <0$, but $x>0$, then we must take $r=-\sqrt{x^2+y^2}$ for the equation to "work out". If the converse is true, we must take the positive value.
In actual usage though, most people prefer to stick with positive $r$, and just change the angle accordingly. Thus instead of using $r=-1, \theta =-\pi$, people commonly just use $r=1, \theta =0$. This means that people just use $r=+\sqrt{x^2+y^2}$,and then choose $\theta$ accordingly.