# How to sketch $\ln\left(\sqrt{x^2+y^2}\right)=-\arctan\left(\frac{y}{x}\right)$ in polar coordinates?

How would one sketch the graph of $$\ln\left(\sqrt{x^2+y^2}\right)=-\arctan\left(\frac{y}{x}\right)$$ in polar coordinates? I'm aware that polar coordinates involve a radius $r$ and angle $\theta$, such as $r=2-\cos(\theta)$, but in this case we have a cartesian equation. How would we convert this in terms of $r$ and $\theta$? Thanks.

• $x^2+y^2 = r^2$ and $\arctan(y/x) = \theta$ Are you familiar with these formulae? Sep 26 '17 at 13:59
• @JohnLou Hi John, what process led to those derivations? (it seems straightforward but I still want to know, thanks!) Sep 26 '17 at 14:02
• The former is derived via the pythagorean theorem (or the distance formula). The same is true of $\theta$. Draw a triangle to help prove this. mathsisfun.com/polar-cartesian-coordinates.html Sep 26 '17 at 14:03
• It's also worth pointing out that $\arctan(y/x)$ is only $\theta$ when $-\pi/2 < \theta < \pi/2$. Sep 26 '17 at 14:03
• So is it just convenient that my equation contains an $x^2+y^2$ and a trig function $\arctan(y/x)$? Otherwise you can't express a Cartesian equation in polar coordinates? Sep 26 '17 at 14:05

$$z=e^{-\theta}e^{i\theta}\\ r=e^{-\theta}$$
$$x=\Re(z)=e^{-\theta}\cos\theta\\ y=\Im(z)=e^{-\theta}\sin\theta\\ x^2+y^2=e^{-2\theta}\\ \sqrt{x^2+y^2}=e^{-\theta}\\ \theta=\tan^{-1}\frac{y}{x}\\$$
$$\ln\left(\sqrt{x^2+y^2}\right)=-\arctan\left(\frac{y}{x}\right)$$