Question: What points in the complex plane satisfies $|z|=arg(z)$.
Work thus far: Let $z=a+bi$ and $arg(z)$ is the angle between the vector $z$ and the axis. So $|z|=\sqrt{a^2+b^2}$ and $a=|z|\cos(\theta),b=|z|\sin(\theta)$. I would expect from intuition that the shape traced out by the points is a spiral. From here I don't how to proceed. Any hints would be appreciated.
Try thinking the other way:
Fix some $\theta_0$ and imagine a line starting at the origin and making an angle $\theta_0$ with the real axis. How many points $z $, in that line, have modulus $|z|$ equal to $\theta_0$? Can you write an expression for it/them?
Let's take a parametric approach: suppose $|z|=\arg(z)=t\ge 0$. Then using polar form and Euler's formula, we can write $z=t\exp(it)$. This traces out the curve
Depending on your perspective, you may choose to only consider $0\le t < 2\pi$.
If you use polar coordinates $z = re^{i\theta}$ so the question is which points are $z = re^{ir}; r\in [0,\infty)$ which is it's own answer.
To convert back to rectangular coordinates we have $z = r\cos r + ir\sin r; r\in [0,\infty)$.
To continue what you started:
$a=|z|\cos(\theta)=|z|\cos(|z|),b=|z|\sin(\theta)= |z|\sin(|z|)$.
As $|z|$ may be any non-negative real value the solution is all $z = r\cos r + ir\sin r; r\in [0,\infty)$
