interpolation with derivative of rational fraction Studying a problem in conformal geometry, I am facing to the following interpolation problem.
Let $P$ and $Q$ two coprime polynomials. Then let $A$ and $B$ two coprime polynomials such that 
$$\frac{A}{B} = \left( \frac{P}{Q}\right)'.$$
We denote by $(z_i)_{0\leq i\leq n}$ some distinct complex numbers with $n=deg(A)$ and which are not some roots of $Q$. Then we consider some perturbations of the $\left( \frac{P}{Q}\right)'(z_i)$ denoted $\tilde{a}_i$ and we try find a perturbation of $P$ and $Q$ denoted $\tilde{P}$ and $\tilde{Q}$ such that
$$\left( \frac{\tilde{P}}{\tilde{Q}}\right)'(z_i)=\tilde{a}_i$$
The problem is easy to solve when $Q\equiv 1$, it is classical interpolation on $P'$ and then we integrate. But When $Q$ is not trivial we can't just perturb $A$, finding $\tilde{A}$ such that 
$$\tilde{A}(z_i)= \tilde{a_i} B(z_i)$$
because when integrating $\frac{\tilde{A}}{B}$ there is no reason to get a rational fraction.
I have tried to perturb successively $P$ and $Q$(or $A$ and $B$) and to get some estimate to make a fixed point but without success? did any one get some references or ideas on that topic?
The Geometric point of view is the following, you consider a (finite) multiple conformal  covering of $S^2$, then identifying $S^2$ with $\hat{\mathbb{C}}$ it is parametrized by a rational fraction. Then you can define the vector field $ \left( \frac{P}{Q}\right)'$. And the question is, if you perturb this vector field at $n+1$ points which are not poles, can you reach this perturbation by perturbing $P$ and $Q$?
Edit: The first question is of course the existence of $\tilde{P}$ and $\tilde{Q}$?
But I would like that $\tilde{P}$ and $\tilde{Q}$ will be closed to $P$ and $Q$ with respect to $\sum \vert a_i -\tilde{a}_i\vert$. Closed means for me, with the same degree and closed in $\mathbb{C}_d[X]$ with $d$ the degree considered.
 A: This is an answer to a slightly different question.
Let $\deg P = p$ and $\deg Q = q$. The polynomials $P$ and $Q$ have $(p+1)+(q+1) = p+q+2$ coefficients between them. However, rescaling $P$ and $Q$ by the same number does not effect the rational function, so there are actually $p+q+1$ parameters available. In addition, if $\deg P \geq \deg Q$ and $a$ is a constant then $(P+a Q)/Q$ is a rational function with numerator and denominator of the same degree, and $(P/Q)' = ((P+aQ)/Q)'$. So there are actually $p+q$ parameters in this case. In short: 

We should expect to be able to interpolate $p+q+1$ points if $p<q$, or
  $p+q$ if $p \geq q$.

The original question asked us to interpolate at $\deg A$ points. We have $\deg A = p+q-1$ if $p \neq q$, or $p+q-2$ if $p=q$ (because the leading terms of $P'Q$ and $P Q'$ cancel.) So the original question is asking for less interpolation than we would expect to be able to achieve. I will show that we cannot always achieve as much interpolation as we would naively expect; I do not know if we can always achieve the smaller amount you ask for.
It is not always possible to perturb an interpolation. The simplest example is to start with $p(x)/q(x) = 1/(x^2+1)$ and $(w_1, w_2, w_3) = (-1, 0, 1)$. So $(a_1, a_2, a_3) = (1/2, 0, -1/2)$. I claim that we cannot perturb to $(1/2, 0, -1/2+\epsilon)$ for $\epsilon \neq 0$, with $\deg p=0$ and $\deg q = 2$. Proof: Such a perturbation is of the form $\phi(x)=1/q(x)$ for some quadratic polynomial $q$. Then $\phi'(x) = -q'(x)/q(x)^2$ and we have imposed that the derivative vanish at $0$. So $q'(x)=0$ and $q(x) = a x^2 + b$ for some $a$ and $b$. This shows that $\phi(x)$ is an even function, so $\phi'(x)$ is odd, and we have $\phi'(1) = - \phi'(-1)$, making it impossible to interpolate $(-1, 0, 1+\epsilon)$.
If you want a counterexample with $p=q$, take $p=q=2$ with $(w_1, w_2, w_3, w_4) = (1, -1, 1/2, 2)$. Try to perturb $(x^2+1)/(x^2+x+1)$ while keeping $a_1=a_2=0$. The condition $\phi'(\pm 1) =0$ implies that $\phi(x) = \phi(1/x)$ so $\phi'(x) = (-1/x^2) \phi'(1/x)$ and we get $\phi'(2) = (-1/4) \phi'(1/2)$, and we can use this to show the impossibility of perturbing to $(0,0,a_3, a_4)$ for generic $(a_3, a_4)$. 
I don't have any counterexamples where the $a_i$ are all nonzero. I also don't have any counter-examples which involve interpolation at only $\deg A$ points.
