Ive asked this question a week ago, but nobody managed to answer but it is doing my heading from then. I know usually You demand some initial work done on the question but I just dont know how to approach it. AS in title $$\log{|z|}=-2\arg(z)$$ In general I guess this could be expressed as $$\log(\sqrt{{x^2+y^2})}=-2\tan^{-1}({\frac{y}{x}})$$ which is hard enough for me to solve. I don't even know if one can do anything to this to separate y's and x's. But even then that would be true only for $z$ in first and fourth quadrant as you add or subtract $\pi$ if z is in second or third quadrant.
Another approach I had was to rewrite it as $$|z|=e^{-2(\arg(z))}$$but againI don't know where to go from here. One solution is definitely z=1 but it is just guesswork. Thank You
Question is taken from the book "Complex Analysis" by Theodore W. Gamelin. I couldn't find a PDF.
Let $\theta=\arg z$ and $r=|z|$. Then the solution set of the equation $\log|z|=-2\arg z$ is the curve whose equation in polar coordinates is $r=e^{-2\theta}$. It is an spiral around the origin. You can get points in cartesian coordinates as $(e^{-2\theta}\cos\theta,e^{-2\theta}\sin\theta)$ for any real $\theta$.