Say I have the equation of a parabola in the complex plane:
$|z-ih| = |z-(x+ib)|$
i. e. the set of all points $z = x+iy$ in the complex plane which are equidistant from the focus $f = ih$ and the line $r = x + i b$, where $b$ and $h$ are real and distinct. Is it then correct to say that $\big|\frac{z-z_0}{e^{i\theta}} -ih\big| = \big|\frac{z-z_0}{e^{i\theta}}-(x+ib)\big|$ is the equation of the parabola I would obtain by rotating the original parabola by an angle $\theta$ and then translating that parabola by $z_0$?
I am using an online complex locus plotter (https://www.geogebra.org/m/FzmbuJd5) and it makes the parabola rotate into a hyperbola. So, am I wrong or this plotter is not quite accurate?
Example of the geogebra plotter with $z_0 = 0$ and $\theta = \pi/2$. The focus is $f = i$ and the directrix is $x-2i$ It plots a hyperbola instead of a rotated parabola.
The original (unrotated) equation works because the plotter treats $x$ as equal to $\Re(z)$ (the real part of $z$) by definition. Hence $z - x = i\Im(z)$ (where $\Im(z)$ is the imaginary part of $z$, which the plotter treats as equal to $y$ by definition) Hence $z-(x+ib)$ works out to $iy - ib$ and $\lvert z-(x+ib)\rvert$ works out to $\lvert y - b\rvert$, the distance from $z$ to the line $y = b$.
The problem seems to be that $x$ in the "rotated" equation is still $\Re(z)$, and subtracting $\Re(z)$ from $ze^{-i\theta}$ does not give you $\Im(z)$ or any other expression that gives you the distance to a directrix line. Instead, it's giving you some factor times the distance to some line, so that you get a hyperbola rather than a parabola, although I have not worked out what line and what factor. (You get a hyperbola when the distance from each point to the directrix is $r$ times the distance to the focus, for some constant $r$ with $0 < r < 1$.)
Here's a formula that seems to work after a fashion, but only for rotation around $0$:
$$ \left\lvert \frac{z}{e^{i\theta}}-ih \right\rvert = \left\lvert y\cos(\theta)-x\sin(\theta) - b \right\rvert. $$
A Geogebra formula (with $\theta = \frac32\pi$ as an example) is
abs(z/exp(i*pi*1.5)-i)=abs(y*cos(pi*1.5)-x*sin(pi*1.5)-2).What I think you want is
$$ \left\lvert \frac{z - z_0}{e^{i\theta}} - ih \right\rvert = \left\lvert \Im\left(\frac{z - z_0}{e^{i\theta}}\right) - b \right\rvert, $$
but I have not figured out how to express that explicitly in Geogebra. An equivalent expression is $$ \left\lvert \frac{z - (p+iq)}{e^{i\theta}} - ih \right\rvert = \left\lvert (y-q)\cos(\theta)-(x-p)\sin(\theta) - b \right\rvert, $$ where $z_0 = p + iq$ for real $p$ and $q$. For $z_0 = 2 + i3$ and $\theta = 1.5\pi$, a Geogebra formula is
abs((z-(2 + i*3)/exp(i*pi*1.5)-i)=abs((y-3)*cos(pi*1.5)-(x-2)*sin(pi*1.5)-1).