It says it all in the title: I need to show how simple complex transformations (translations and dilations, or even both) affect shapes on the complex plane in a "fluid" way – that is, creating some kind of animation that shows how the transformation works by continuously bending/gliding an object from its original shape to its image. I'm working on Desmos, which offers the possibility of creating sliders (much like GeoGebra), which work like this.
Translations are very simple. Starting from a complex number $x+iy$, translation by $a+ib$ (with $x,y,a,b \in \mathbb{R}$) means
$$\mathcal{T}(z) = (x+a) + i(y+b)$$
I can then add the glider $k$ to get
$$\mathcal{T}(z) = (x+ka) + i(y+kb), \quad 0 \le k \le 1$$
It's clear that when $k=0$ we have the identity transformation, and when $k=1$ the translation is over.
I'm having a hard time depicting dilations in the same way. Dilations by $a+ib$ take the form
$$\mathcal{D}(x+iy) = (ax-by) + i(bx+ay)$$
My thought was that I need to place $k$ in such a way that when $k=0$ the expression reverts to $x+iy$, and when $k=1$ the expression looks like the equation above. So I conjectured that this would work:
$$\mathcal{D}(x+iy) = (a^kx-kb^ky) + i(kb^kx+a^ky), \quad 0\le k\le 1$$
Yet the formula blows up whenever either $a$ or $b$ or both are negative, for non-natural $k$. Is there a nicer and more effective way to insert a slider in the dilation equation above?
Addendum. The objective would be having the means of depicting not just simple transformations but also any other function that can be found out there, given its real and imaginary parts. I'm looking especially at $f(z) = z^{-1}$, which would allow me to represent Möbius transformations.
Dilation by $1$ is the identity function, so given a path from $1$ to $z = a+ib$ you obtain a continuous transformation from identity function to dilation by $z$. However, this transformation can look strange or ugly, depending on the path chosen.
Since complex multiplication is, geometrically, a scaling and a rotation, I would recommend to put $z$ into polar form $z = re^{i\varphi}$ and create a path from $1$ to $z$ by $\gamma(t) = tre^{it\varphi}$, and the final transformation looks like $(x+iy) tr e^{it\varphi}$. This way you get a continuous monotone scaling and rotation.
Unfortunatelly, I guess you cannot do such a transformation for $z^{-1}$ without passing some degenerate function in the middle, basically because $z^{-1}$ changes the orientation of the plane.