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Im trying to find all set of points on the complex plane for which $|z|=\arg z$.

I rewrote $|z|= \sqrt{x^2 + y^2}$ and $\arg z$ as $\tan^{-1}(y/x)$.I set them equal But im not sure what to do next.

StubbornAtom
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Eamonn
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  • When you set (when $\text{z}\in\mathbb{C}$):

    $$\arg\left(\text{z}\right)=\arg\left(\Re\left[\text{z}\right]+\Im\left[\text{z}\right]i\right)=\arctan\left(\frac{\Im\left[\text{z}\right]}{\Re\left[\text{z}\right]}\right)$$

    That is only true when $\Re\left[\text{z}\right]>0$ and $\Im\left[\text{z}\right]>0$.

     – Jan Eerland Oct 03 '16 at 17:52

2 Answers2

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Interpret ${\rm arg}$ as principal value ${\rm Arg}$, and write $z$ in the form $$z=r(\cos\theta+i \sin\theta)\qquad(r\geq0,\quad -\pi<\theta<\pi)\ .$$ Your condition then amounts to $$r=\theta\ ,$$ so that necessarily $r=\theta>0$, since $\theta$ is undefined at $z=0$. This shows that the set $$S:=\bigl\{z=\theta\,(\cos\theta+i\sin\theta)\in{\mathbb C}\>\bigm|0< \theta<\pi\bigr\}\ ,$$ an arc enter image description here of an Archimedean spiral enter image description here

is the solution to your problem.

MarianD
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Christian Blatter
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Let $|z|=\arg z=t$. Then by definition

$$z=|z|e^{i\arg z}=te^{it},$$

(in Cartesian coordinates $x=t\cos t,y=t\sin t$) which is an Archimedean spiral .