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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.

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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 $\theta$.

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  • $\begingroup$ Thank you but it was more of an answer (which i wanted) rather than explanation. Can you explain to me or give me a reference to where "Then" comes from ? $\endgroup$ – Scavenger23 Feb 5 '16 at 16:36
  • $\begingroup$ Seems the "then" comes from just taking $e$ to power both sides. $\endgroup$ – peter.petrov Feb 5 '16 at 16:38
  • $\begingroup$ Consider the ray $\arg z=\theta$. On it here is a unique point $z$ with $|\logz|=-2\theta$, or $|z|=e^{-2\arg z}$. $\endgroup$ – Julián Aguirre Feb 5 '16 at 16:40
  • $\begingroup$ I had let $z=re^{i\theta}$ so that $\log|z|=\log(r)$ and $-2\arg(z)=-2\theta$ $\Rightarrow$ $\log(r)=-2\theta$ $\Rightarrow$ $10^{-2\theta}=r$. What am I doing wrong? Or does $\log$ mean the natural logrithm $\ln$? $\endgroup$ – TheLast Cipher Jul 1 '18 at 2:19
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    $\begingroup$ Yes, $\log$ is the natural logarithm. $\endgroup$ – Julián Aguirre Jul 1 '18 at 6:53

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