TITLE: Sketch $\log(|z|)= -2\arg(z)$
QUESTION [1 upvotes]: 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

REPLY [1 votes]: 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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