Let's have $a_0$, $a_1$, $a_2$ complex numbers which are on the circle $(x-2)^2 +y^2=1$ (so for all of them the abs$\leq 3$ $\left|a_0\right|\leq3$, $\left|a_1\right|\leq3$, $\left|a_2\right|\leq3$). Let's have $v$ another complex number, and suppose that $v^3+a_2v^2+a_1v+a_0=0$. We must prove that $\left|v\right|<4$. The mathematical solusion is easy: $$\left|v^3\right|\leq3(\left|v\right|^2+\left|v\right|+1)\implies\left|v^3\right|-1\leq3(\left|v\right|^2+\left|v\right|+1)-1\implies\dots\implies\left|v\right|-1<3\\ \implies\left|v\right|<4$$ I'm asking if there is a way to plot the curve $v^3+a_2v^2+a_1v+a_0=0$ and show with the plot that $\left|v\right|<4$ in all cases. Thanks for your time

  • $\begingroup$ What do you mean by the curve v^3+a2*v^2+a1*v+a0 = 0 ? To plot all the three curves (three roots of the equation depending on a0, a1, a2)? If so then it should be a three-dimensional submanifold, not just one-parameter curve. $\endgroup$
    – Artes
    Jun 3 '13 at 16:10
  • 3
    $\begingroup$ @artes When I went to school we actually hand-plotted things when doing math. So plotting doesn't necessarily imply Mathematica. But we may give the OP the benefit of doubt. I'd appreciate some kind of acknowledgment by the OP that this question is indeed to be answered in the context of the Mathematica program. $\endgroup$ Jun 3 '13 at 16:43

Since there is no straightforward approach to plotting "curves" depending on three parameters one can use Manipulate with three controls. We have three complex numbers a0, a1, a2 but since they lie on an appropriate circle we can parametrize them by three real parameters u1, u2, u3. Next we have to express solutions to the equation v^3 + a2 v^2 + a1 v + a0 == 0, we can do it with
Root[#^3 + (2 + Exp[I u3]) #^2 + (2 + Exp[I u2]) # + (2 + Exp[I u1]) &, 2]] where we substitute a0 by 2 + Exp[I u1], a1 by 2 + Exp[I u2] etc.

Now we have

     ContourPlot[{ (x - 2)^2 + y^2 - 1 == 0, x^2 + y^2 - 16 == 0},
                 {x, -4, 4}, {y, -4, 4}, Axes -> True, 
                 ContourStyle -> {{Thick, Blue}, {Thick, Darker@Green}}, 
                 AxesStyle -> Arrowheads[0.07]], 

       Table[{k[[1]], PointSize[0.02],
                      Point[{2 + Re[Exp[I k[[2]]]], Im[Exp[I k[[2]] ]]}]},

             {k, {{Red, u1}, {Darker@Green, u2}, {Darker@Orange, u3}}}], 
       Table[{k[[1]], PointSize[0.03], Point[{Re@#, Im@#} &@
              Root[#^3 + (2 + Exp[I u3]) #^2 + (2 + Exp[I u2]) # + (2 + Exp[I u1]) &,
             {k, {{Blue, 1}, {Magenta, 2}, {Darker@Cyan, 3}}}]}]], 

              {u1, 0, 2 Pi}, {u2, 0, 2 Pi}, {u3, 0, 2 Pi}]

enter image description here

Now this becomes evident that all the solutions represented by blue, magenta and cyan points satisfy Abs[v] < 4, in other words they lie in the green circle.

  • $\begingroup$ Thank you very much for your help. $\endgroup$
    – user81175
    Jun 6 '13 at 5:11
  • $\begingroup$ @lafkasd If you've asked this question consider registering your account and merge it with user7851. It hasn't been obvious if this question regarded Mathematica programming or you just plotting certain mathematical object (so how to plot it ?). Therefore they have sent it here. $\endgroup$
    – Artes
    Jun 6 '13 at 9:24
  • $\begingroup$ Could this method verify that |v| cannot actually exceed 3? This is merely conjectured (with proof for |v| < 3.0154). $\endgroup$
    – user120385
    Jan 10 '14 at 11:27

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