Your problem is:
f = @(z) (0.10-0.3i)*z.^(-1) + (0.2121 - 0.0008i)*z.^(-2) +(0.9+0.001i)*z.^(-3);
Please don't use curly braces in an equation. That is not valid MATLAB syntax. As well, MATLAB uses * to multiply. There is no implicit multiplication by a constant. MATLAB finds the solutions as:
zsol = solve(f(Z),Z,'maxdegree',3)
zsol =
The solution you claim to have found using MATLAB is a spurious one, since z == 0 is a divide by zero. So z = 1/0 is NOT a solution to that equation, even though you find it by substituting z = 1/u, and then solving for the roots using u. Sorry. There are only TWO solutions.
vpa(zsol)
ans =
And what are the inverses ofthose numbers, so we can compare it to what you claim to have found using MATLAB?
1./vpa(zsol)
ans =
And that is the same thing that MATLAB returns.
We can verify these are indeed the true solutions to the problem you show.
f(vpa(zsol))
ans =
So numbers on the order of 1e-40. Since this is the default number of digits for vpa, that is just floating point trash.
Is there any meaningful reason to perform this solution a million times, or even just 100? NO. These are the solutions, and they don't change with time.
Anyway, are the solutions you claim to have found by hand really solutions to the problem you show?
z2invclaimed = -0.4929597603-0.4990725263i;
So not in fact a solution. It is close. You have made approximations, truncating results to only a few significant digits, so your claimed to be solutions that you found by hand are only approximate solutions. Effectively, they are solutions to approximately the problem you have written down.
Again, repeating the solution as many times as you wish will not matter. The result will not change.
