To get you started:
1. Define:
x_{0}=2\pi{10}^{6}
x_{1}=2\pi{10}^{10}
Ask yourself:
Why have you been given so huge arguments?
In particular, can I use that fact to my advantage later on?
2.Rewrite:
f(x)=a+\frac{b-a}{1+jxc}=a+\frac{1-jxc}{1-jxc}\frac{b-a}{1+jxc}=\frac{ax^{2}c^{2}+b}{1+x^{2}c^{2}}+j\frac{(a-b)xc}{1+x^{2}c^{2}}
3. Requirements of curve fitting:
\frac{ax_{0}^{2}c^{2}+b}{1+x_{0}^{2}c^{2}}\approx{2.33}
\frac{ax_{1}^{2}c^{2}+b}{1+x_{1}^{2}c^{2}}\approx{2.347}
\frac{(a-b)x_{0}c}{1+x_{0}^{2}c^{2}}\approx{-1.16510*10^{-3}}
\frac{(a-b)x_{1}c}{1+x_{1}^{2}c^{2}}\approx{-3.755210*10^{-3}}
4. Define:
y_{0r}=2.33,y_{1r}=2.347,y_{0i}=-1.16510*10^{-3},y_{1i}=-3.755210*10^{-3}
5. Define:
\hat{y}_{0r}=\frac{ax_{0}^{2}c^{2}+b}{1+x_{0}^{2}c^{2}}
\hat{y}_{1r}=\frac{ax_{1}^{2}c^{2}+b}{1+x_{1}^{2}c^{2}}
\hat{y}_{0i}=\frac{(a-b)x_{0}c}{1+x_{0}^{2}c^{2}}
\hat{y}_{1i}=\frac{(a-b)x_{1}c}{1+x_{1}^{2}c^{2}}
6. Construct:
S(a,b,c)=(y_{0r}-\hat{y}_{0r})^{2}+(y_{0i}-\hat{y}_{0i})^{2}+(y_{1r}-\hat{y}_{1r})^{2}+(y_{1i}-\hat{y}_{1i})^{2}
Clearly, S>=0, and S=0 if and only if the curve fitting is exact.
We are interested in the choice of (a,b,c) such that a minimum of S is found.
Hence, we should consider the system of 3 equations:
\frac{\partial{S}}{\partial{a}}=0
\frac{\partial{S}}{\partial{b}}=0
\frac{\partial{S}}{\partial{c}}=0
This system can (theoretically, at least!) be solved for minimizing values
(a_{m},b_{m},c_{m})
To find a simple, approximate solution to the system of equations, I suggest that you utilize your knowledge that (x_{0},x_{1}) are huge numbers.
Good luck!
NOTE:
This is just one of many techniques to derive curve-fitting coefficients.
It is by no means clear that this technique provides the simplest system to solve for coefficients (a,b,c). Look up in a numerical analysis book (or something like that) to get other ideas..
