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I believe I can solve for z:

Let $z=re^{i\theta}$, then $\log(|z|)=-2\arg(z)$ becomes $$\log(r)=-2\theta$$ ->

$$z=e^{-2\theta}[\cos(\theta)+i\sin(\theta)], 0<\theta \leq \pi$$

But I don't know how to use this fact to sketch the whole set of solutions. Since both $\log(|z|)$ and $\arg(z)$ are real-valued, I tried to do it like this:

For $x>0$ $$\log(\sqrt{x^2+y^2})=-2\Bigl(\tan{\frac{y}{x}}\Bigr)^{-1}$$

-> $$x^2+y^2=e^{-4\tan\left(\tfrac{y}{x}\right)^{-1}}$$

but I have no idea how to proceed. And this is only the case when $x>0$. Thanks for any help

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With $z=re^{i\theta}$ the relation $$\log |z|=-2\arg z$$ will be $r=e^{-2\theta}$ which shows a spiral shape named logarithmic spiral.

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  • $\begingroup$ yes, thats what I meant in my answer, I lost the "-" sign. but if I never met the logarithmic spiral, how do I know the answer. and moreover how do I sketch the region in the cartesian plane? $\endgroup$ – Scavenger23 Nov 15 '17 at 19:56
  • $\begingroup$ See the link. With some values for $\theta$ you can draw it's shape in every interval. $\endgroup$ – Nosrati Nov 15 '17 at 19:58
  • $\begingroup$ wolframalpha.com/input/… $\endgroup$ – Nosrati Nov 15 '17 at 20:02

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