The complex inquality $|z-2+3i| < 5 , z\in \mathbb{C}$ has to be resolved and geometrically interpreted. I am stuck with this:
Set $z=a+bi$ and $|a+bi-2+3i|<5 \\ \sqrt((a-2)^2+(b+3)^3)<5 \\ a^2+b^2-2a+6b<12$
I plotted it, but even with the plot I can't help myself. I am not able to derive this parabolic form.
On
$$|z-2+3i|<5\iff |z-(2-3i)|<5$$This describes an open disk in the complex plane centered at $z_0=2-3i$ with a radius of $5$. This is because the set $\{z\in\Bbb C:|z-(2-3i)|<5\}=\{z\in\Bbb C:|z-z_0|<5\}$ is all points in the complex plane such that the distance to the point $z_0=2-3i$ is less than $5$, which is precisely a disk of radius $5$.
Write it as $$|z-(2-3i)|<5$$ Then the left side is just the distance between the point $z$ and the point $2-3i$.
This means the points satisfying the inequality are precisely the points which are less than $5$ units away from $2-3i$, which are the points in the open disc of radius $5$ centered at $2-3i$.
Generally speaking, if $r$ is a positive number, then
$|z-z_0|<r$ represents an open disc of radius $r$ centered at $z_0$
$|z-z_0|\leq r$ represents the corresponding closed disc
$|z-z_0|=r$ represents the circle (the boundary of the disc)
$|z-z_0|>r$ represents the (open) exterior of the disc (excluding the bounding circle)
$|z-z_0|\geq r$ represents the (closed) exterior of the disc (including the bounding circle).