Moving plane over a fixed plane I am reading the following work
https://www.mathcurve.com/courbes2d.gb/base/base.shtml
and I need some explanation about the terminology.


*

*There is a problem with respect to the definition of the point $B(t)$
of affix $b(t)$. It is said that it's the point of the base which should satisfy : $b'(t)=\mu'(t)=0$. But if we take the example of a circle rolling of the plane (treated a bit below in the same page), that is the base is the set of points of affix $b(t)=t$ and $\forall t,\; b'(t)\neq 0$. But it is well known that the instant centre of rotation for this particular problem is the point of contact between the circle and the line. What is the problem?

*What does the moving plane and fixed plane corresponds to ?
The fixed plane would be the surface containing the base curve and orthogonal to your screen, but it is not a plane, it is a surface.
Otherwise it seems that by fixed plane, he means the fixed reference frame $(I,i,j)$?


*How to prove that $d b = \omega+e^{i\theta} d\beta$ ? Why does it correspond to the non slipping case ?

*Regarding the equation $dt = e^{i\theta}\,i\,R\,e^{iu}\, du$, why do we obtain: $e^{i\theta}=-ie^{-iu}$.

 A: The problem comes from the interpretation of the parametrized base curve and the rolling curve in terms of the time parameter $t$ and there is a confusion about what exactly you are parametrizing and tracking.   
To visualize this whole thing, picture the moving plane as a transparency slide for overhead projector (the transparent slides on which you can write stuff with a marker and then project on a screen when you give a talk), while the fixed plane as a piece of white paper underneath. The rigid motion of one plane over the other is basically the sliding of the transparency slide (moving) over the fixed white paper underneath. To be able t track where points go better, draw two coordinate systems with a marker: 
(a) coordinate system $\Omega\, \vec{k}\vec{l}$ on the transparency slide (this is the moving coordinate system)
(b) coordinate system $O\, \vec{i}\vec{j}$ on the white paper under the slide (this is the fixed coordinate system)
At first, at time $t=0$, if you wish, you can draw both coordinate systems to overlap.
I suppose you are ok with the usage of complex numbers as vectors and rotations in the plane. You can think of the axis $\Omega\vec{k}$ as the real axis of the slide, seen a copy of the plane of complex numbers, and  $\Omega\vec{l}$ as the imaginary axis. Analogously,  $O \vec{i}$ is the real axis of the white paper under the slide and  $O\vec{j}$ is the imaginary axis. As time passes, the location of the point $\Omega$ on the white paper changes, while on the slide it is always represented by $0$. Furthermore, the basis vectors $\vec{k}$ and $\vec{l}$ drawn on the slide, i.e. fixed on the slide, change orientation with time with respect to the paper, i.e. with respect to $O\, \vec{i}\vec{j}$. The rigid motion of the slide over the white paper can be described by simply tracking the position of $\Omega$ and the orientation of the vectors $\vec{k}$ and $\vec{l}$ with time as seen from the white paper's coordinate system $O\,\vec{i}\vec{j}$. Thus, 
a) the complex number, representing the position vector of $\Omega$ in $O\,\vec{i}\vec{j}$, is denoted by $\omega(t)$;
b) the time-dependent orientation of the unit vector $\vec{k}$ with respect to $O\,\vec{i}\vec{j}$ can be represented by the unit complex number $e^{i\, \theta(t)}$, interpreted as both a unit vector and as a rotation of the number $1$.
c) the time-dependnt orientation of the unit vector $\vec(l)$ in $O\,\vec{i}\vec{j}$  is $i\, e^{i\, \theta(t)}$, where $i$ here is the imaginary unit and not the vector $\vec{i}$! you can think of this again as the rotation of the number $i$.
If you draw an arbitrary point $M$ on the transparency slide, it is fixed on the slide but moves with respect to the white paper, i.e. $M$ is fixed with respect to $\Omega\vec{k}\vec{l}$ and it moves with respect to $O\,\vec{i}\vec{j}$. The position vector of a point M on the slide, which in our case is represented by a time-independent complex number, is $\mu$ with respect to $\Omega \,\vec{k}\vec{l}$. At time $t$, the position of $M$ with respect to the white paper, i.e. with respect to the coordinate system $O\,\vec{i}\vec{j}$, is the vector (complex number) $m(t)$.
To represent the vector $\vec{\Omega M}$ at time t in the basis $\vec{i}\vec{j}$, we observe that $\vec{\Omega M} = \vec{O M} - \vec{O \Omega} = m(t) - \omega(t)$. At the same time, the same vector $\vec{\Omega M}$ in the basis $\vec{k}\vec{l}$ is given as 
$\vec{\Omega M } = \mu =\mu_1 + i\, \mu_2 = \mu_1 \, \vec{k} + \mu_2 \, \vec{l}$. However, the basis $\vec{k}\vec{l}$ can be written in terms of the basis $\vec{i}\vec{j}$ where $\vec{k} = e^{i\theta(t)}$ and $\vec{l} = i\,e^{i\theta(t)}$. Therefore in the basis paper basis $\vec{i}\vec{j}$ the vector
$\vec{\Omega M }   = \mu_1 \, \vec{k} + \mu_2 \, \vec{l} = \mu_1\, e^{i\theta(t)} + \mu_2\, i\, e^{i\theta(t)} = e^{i\theta(t)}\big(\mu_1 + i\, \mu_2\big) = e^{i\theta(t)} \mu$. Thus $m(t) - \omega(t) = \vec{\Omega M}= e^{i\theta(t)} \mu$ so $m(t) = \omega(t) + e^{i\theta(t)}\mu$ is the change of coordinates from the point of view of the slide $\mu$ to the point of view of the paper $m(t)$. 
As we can see from $m(t) = \omega(t) + e^{i\theta(t)} \mu$ that the rigid motion of the slide over (with respect to) the white paper is a composition of a rotation and translation. As such, at any moment of time $t = \tau$, this rigid motion is equivalent to a rotation around some point. So at time $\tau$ we are looking for a point $B_s(\tau)$ on the slide, and its twin point, we call it $B_p(\tau)$, on the paper (at time $\tau$ the two points, $B_s(\tau)$ drawn with marker on the slide and $B_p(\tau)$ drawn with marker on the paper coincide, i.e. overlap, but for other moments of time they are different), that represent the point of rotation  on the slide and on the paper respectively of the rigid motion $m(t) = \omega(t) + e^{i\theta(t)}\mu$ at the time moment $t=\tau$. The point $B_s(\tau)$ with respect to the slide has complex coordinate $\beta_{\tau}$. This coordinate is not time-dependent! The complex coordinate of $B_s$ with respect to the paper at any time $t$ is $b_s(t,\tau) = \omega(t) + e^{i\theta(t)}\beta_{\tau}$. One can show that $B_s$ is the momentary fixed point for the rigid motion of the slide relative to the paper at the moment of time $t = \tau$ exactly when $\frac{db_s}{dt}(\tau, \tau) = \dot{b}_s(\tau,\tau) = 0$. However, for other times $t \neq \tau$ generally $\dot{b}_s(t,\tau) \neq 0$. 
To find $B_s(\tau)$, one can show that the $B_s(\tau)$ is the momentary fixed point of the rigid motion $m(t) = \omega(t) + e^{i\theta(t)}\mu$ when $t=\tau$ if and only if $\frac{db}{dt}(\tau) = \dot{b}(\tau) = 0$, i.e. $$0 = \frac{d\omega}{dt}(\tau) + \Big(\,\frac{d}{dt}e^{i\theta(t)}\,\Big)\big|_{t=\tau}\beta_{\tau} = \dot{\omega}(\tau) + i\, \dot{\theta(\tau)}\, e^{i\theta(\tau)}\beta_{\tau} = \dot{\omega} + i\, \dot{\theta}\, e^{i\theta}\beta_{\tau}$$ so when you solve this for $\beta_{\tau}$
$$\beta_{\tau} = \beta(\tau) = i\frac{\dot{\omega}(\tau)}{\dot{\theta}(\tau)}e^{-\,i\theta(\tau)}$$ Now, what this equations tells you that on the slide, with respect to the coordinate system $\Omega\, \vec{k}\vec{l}$, the momentary point of rotation at time $t=\tau$ has a complex coordinate
$\beta(\tau) = i\frac{\dot{\omega}(\tau)}{\dot{\theta}(\tau)}e^{-i\theta(\tau)}$. For different times $t\neq\tau$ the point $B_s(t) \neq B_s(\tau)$ on the slide is different! It changes, it is not the same point! That is why $\beta(\tau)$ changes with $\tau$, it parametrizes in terms of time moments $\tau$ all points on the slide that turn into a momentary point of rotation at time $t = \tau$ but in general for other times the point $B_s(\tau)$ with $\beta(\tau)$ is not the momentary fixed point of rotation. This set of momentary rotation points, as seen from the paper's coordinate system $O\, \vec{i}\vec{j}$, is given by the change of complex coordinate
$$b(\tau) = \omega(\tau) + e^{i\theta(\tau)}\beta(\tau) = \omega(\tau) +  e^{i\theta(\tau)}i\frac{\dot{\omega}(\tau)}{\dot{\theta}(\tau)}e^{-i\theta(\tau)} = \omega(\tau) +  i\frac{\dot{\omega}(\tau)}{\dot{\theta}(\tau)}$$ This equation parametrizes all points $B_p(\tau)$ that become momentary rotation points on the paper at time $t = \tau$ but for other times are generally not momentary rotation points. This is not the equation of rigid potion and is not supposed to have $\dot{b}(\tau) = 0$! 
Thus, the equation of the base is $$ b(\tau) = \omega(\tau) +  i\frac{\dot{\omega}(\tau)}{\dot{\theta}(\tau)}$$ and it is drawn on the paper, i.e. with respect to $O\, \vec{i}\vec{j}$,
while the equation of the rolling curve is $$\beta(\tau) = i\,\frac{\dot{\omega}(\tau)}{\dot{\theta}(\tau)}e^{- i\tau(\tau)}$$ and it is drawn on the transparency slide, i.e. with respect to $\Omega\, \vec{k}\vec{l}$. 
Now, 
\begin{align}
\dot{b}(\tau) =& \frac{db}{d\tau}(\tau) = \frac{d}{d\tau} \left(\omega(\tau) + e^{i\theta(\tau)}\beta(\tau)\right) =  \frac{d\omega}{d\tau}(\tau) + i\,\frac{d\theta}{d\tau}(\tau)\, e^{i\theta(\tau)}\beta(\tau) + e^{i\theta(\tau)}\frac{d\beta}{d\tau}(\tau)\\
=&  \, \dot{\omega}(\tau) + i\,\dot{\theta}(\tau)\, e^{i\theta(\tau)}\beta(\tau) + e^{i\theta(\tau)}\dot{\beta}(\tau) \, = \, \dot{\omega}(\tau) + i\,\dot{\theta}(\tau)\, e^{i\theta(\tau)}\left(i\,\frac{\dot{\omega}(\tau)}{\dot{\theta}(\tau)}e^{- i\theta(\tau)}\right)+ e^{i\theta(\tau)}\dot{\beta}(\tau) \\
=&  \, \dot{\omega}(\tau) + i^2\,e^{i\theta(\tau)} e^{- i\theta(\tau)}\left(\dot{\theta}(\tau) \, \frac{\dot{\omega}(\tau)}{\dot{\theta}(\tau)}\right) + e^{i\theta(\tau)}\dot{\beta}(\tau) \, = \,   \dot{\omega}(\tau) -  \dot{\omega}(\tau) + e^{i\theta(\tau)}\dot{\beta}(\tau)\\
=& \, e^{i\theta(\tau)}\dot{\beta}(\tau)
\end{align}
Thus
$$\frac{db}{d\tau}(\tau) = e^{i\theta(\tau)}\frac{d\beta}{d\tau}(\tau)$$
Therefore, the length of the arc on the base curve that has been in contact with the rolling curve between any arbitrary moments $\tau_1 < \tau_2$ is
\begin{align}
\text{length of base curve between}(\tau_1,\tau_2) =& \int_{\tau_1}^{\tau_2} \left|\,\frac{db}{d\tau}(\tau)\,\right| \, d\tau = \int_{\tau_1}^{\tau_2} \left|\,e^{i\theta(\tau)}\frac{d\beta}{d\tau}(\tau)\,\right| \, d\tau\\
=&  \int_{\tau_1}^{\tau_2} \left|\,e^{i\theta(\tau)}\,\right| \, \left|\,\frac{d\beta}{d\tau}(\tau)\,\right| \, d\tau\\ =&  \int_{\tau_1}^{\tau_2} \, \left|\,\frac{d\beta}{d\tau}(\tau)\,\right| \, d\tau \\
=& \text{ length of rolling curve between}(\tau_1,\tau_2) 
\end{align}
so the lengths of both arcs are equal for any two $\tau_1 < \tau_2$ which means there is no slipping but strict isometric correspondence between the points of both curves.
An finally, to understand the rest, I will show how one can recover the formula for the rigid motion $m(t) = \omega(t) + e^{i\theta(t)}\mu$ (this is the formula for the roulettes depending on the initial condition $\mu$), given the (time paramwetrized) formulas for the base curve $b(t)$ and the rolling curve $\beta(t)$.
First, differentiate $$\dot{b}(t) = \frac{db}{dt}(t) \, \text{ and } \, \dot{\beta}(t) = \frac{d\beta}{dt}(t)$$ Then, from the formula above
$$\dot{b}(t) = \frac{db}{dt}(t) = e^{i\theta(t)}\frac{d\beta}{dt}(t) = e^{i\theta(t)}\dot{\beta}(t)$$ we find
$$e^{i\theta(t)} = \frac{\dot{b}(t)}{\dot{\beta}(t)}$$ From the coordinate transformation between the two curves
$$b(t) = \omega(t) + e^{i\theta(t)}\beta(t)$$ we find
$$\omega(t) = b(t) - e^{i\theta(t)}\beta(t) = b(t) -\frac{\dot{b}(t)}{\dot{\beta}(t)}\beta(t)$$
Thus, the formula for the rigid motion is
$$m(t) = b(t) -\frac{\dot{b}(t)}{\dot{\beta}(t)}\beta(t) \, + \, \frac{\dot{b}(t)}{\dot{\beta}(t)}\mu$$
In the case of a circle of radius $R$ rolling uniformly without slipping along a straight line, without loss of generality, we can assume that 
$\beta(t) = R e^{it}$. Then the length of the straight line traversed by the circle from time $0$ to time $t$ is equal by the arc that has been in touch with the line in the same time interval, which is the central angle of rotation $t$ of the circle times radius $R$, i.e. the line should be parametrized as $b(t) = R t$. The derivatives are
$$\dot{b}(t) = R \, \text{ and } \, \dot{\beta}(t) = i\,R e^{ it}$$  Then
$$\frac{\dot{b}(t)}{\dot{\beta}(t)} = \frac{R}{  i\,R e^{ it}} = - i e^{- it} $$
Thus, the equation of the rigid motion, i.e, the motion of each point $\mu$ on the transparency slide (on which  the circle is drawn) with respect to the underlying sheet of paper is 
$$m(t) = R t + i e^{- it} R e^{ it} - i e^{ - it} \mu =  R t + i R - i e^{- it}\, \mu $$
In particular, if $\mu = R$, then
$$m(t) =  R\, t + i R - i R e^{- it}  = R \big(\, t + i - i(cos(t) - i\sin(t))\,\big) = R\big(\, t - \sin(t) + i(1 - \cos(t) )\,\big)$$ which are the equations of the cycloid. 
