Complex roots of sin(2*x)-2*x=0

How can i use fsolve to obtain the complex roots of the equation: sin(2*x)-2*x=0?

[Matt J]

Well, you definitely can't find more than one. fsolve is a numerical solver.

Answer 1

[Star Strider]

Providing fsolve with a complex initial estimate encourages it to find complex roots —

f = @(x) sin(2*x)-2*x;
xrts = fsolve(f, 1+1i)

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

xrts = 0.0320 + 0.0197i

.

[Matt J]

But I think this is just an approximately converged neighbor of the trivial root at x=0.

f=@(a,b) abs(  sin(2*(a+1i*b))-2*(a+1i*b));
fsurf(f,[-0.1 +0.1, -0.1, 0.1]); view(34,55)

[Star Strider]

True.

However the request was to how to return a complex root. We know nothing more about the intended problem.

.

[Matt J]

Providing fsolve with a complex initial estimate encourages it to find complex roots

Only if the objective is analytic, see

https://www.mathworks.com/help/optim/ug/complex-numbers-in-optimization-toolbox-solvers.html

It's not clear to me whether that is true or not for sin(2*x)-2*x.

Answer 2

[Matt J]

This seems to find a non-trivial complex root:

opts=optimoptions('fsolve','StepTol',1e-14,'FunctionTol',1e-14,'OptimalityTol',1e-14);
[p,fval]=fsolve(@eqnfun,[3,3],opts);

Equation solved, inaccuracy possible.

The vector of function values is near zero, as measured by the value
of the function tolerance. However, the last step was ineffective.

x=complex(p(1), p(2)), 

x = 3.7488 + 1.3843i

sin(2*x)-2*x

ans = 0.0000e+00 + 2.2204e-15i

function F=eqnfun(p)

  x=complex(p(1), p(2));
  y=sin(2*x)-2*x;
  F=[real(y); imag(y)];

end

[Matt J]

There is obviously also a solution at x = -3.7488 - 1.3843i

[Saeid]

Thanks Matt. Just in case you have asked yourselves how an equation like this occurs: in solving certain biharmonic equations the solution at some point requires obtaining the eigenvalues (pN) that are complex roots of this equation:

sin(2pN)=2pN

So, not only is there a complex solution to this equation, but there is apparently an infinite number of solutions