0
$\begingroup$
Z = 1 + I; z0 = Re[Z]; z1 = Im[Z]; 

h[x0_, x1_] := Module[{u = x0, v = x1}, 
  f[z_] := z;
  y[x_] := ((z1 - v)/(z0 - u)) (x - u) + v; 
  u1 = u; 
  w = ComplexExpand[f[x + I y[x]]]; 
  p1 = Manipulate[Plot[y[x], {x, u1, r}, PlotRange -> {{-4, 4}, {-4, 4}},  
          PlotStyle -> {Red, Thick}],{r, 0.1 + u1, z0}]; 
  p2 = If[TrueQ[Element[w, Complexes]],
    ListPlot[{Re[w], Im[w]}, PlotRange -> {{-2, 1}, {-1, 1}},PlotStyle -> {Green,PointSize[0.05]}], 
    Manipulate[ParametricPlot[{Re[w], Im[w]}, {x, u1, r},PlotRange -> {{-4, 4}, {-4, 4}},PlotStyle -> {Green, Thick}], {r, 0.1 + u1, z0}]]; 

   GraphicsGrid[{{p1, p2}}]
];


Manipulate[h[p, q], {p, z0 - 1, z0 + 1}, {q, z1 - 1, z1+ 1}]

I am trying to solve the following problem (or similar to other same type of problems) $$\lim\limits_{z\rightarrow 1+i} z = 1+i.$$

For this I am trying to check that $f[z]\rightarrow 1+i$ through all linear path $z\rightarrow 1+i$. The above program is working but inner Manipulate command is not working..

Kindly, help to solve the problem and if possible then improve the above program for all possible paths $z$ which approaches to $1+i$.

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  • $\begingroup$ The more compact example the more likely you will attract attention. Are all those complex functions needed to reproduce the problem, wouldn't Sin do? You could get rid of PlotStyle/PlotRange etc etc. $\endgroup$ – Kuba Feb 3 at 12:00

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