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

Question

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.

Answer 1

This appears to be a logarithmic spiral. In the complex plane we have

$$z=e^{-\theta}e^{i\theta}\\ r=e^{-\theta}$$

Thus,

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

So that

$$\ln\left(\sqrt{x^2+y^2}\right)=-\arctan\left(\frac{y}{x}\right)$$