Is it possible to use the "fsolve" function in the Optimization Toolbox to solve complex equations?

A function of a complex variable "z" is called a complex function "f(z)". I have the following complex function:

f(z) = z^2 - z + 1

How do I determine the zeros of this complex function?

Answer 1

[MathWorks Support Team]

For release R2016b and after: 

Generally, Optimization Toolbox solvers do not accept or handle objective functions or constraints with complex values. However, the least-squares solvers "lsqcurvefit", "lsqnonlin", and "lsqlin", and the "fsolve" solver can handle these objective functions with some restrictions as mentioned in the following documentation:

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

The problem statement can be solved as follows:

>> f = @(z)1 + z^2 - z;
>> x0 = 1 + 1i/10;
>> y = fsolve(f,x0)

For release R2016a and before:

The "fsolve" function cannot directly handle complex-valued functions or variables. However, a complex-valued function of a complex variable "z" can also be thought of as a system of 2 real-valued equations in 2 real variables.

Consider the problem statement:

f(z) = z^2 - z + 1

To solve this complex function using the "fsolve" function, we first create the following function file:

function F = myfun(X)
    z = X(1, :) + j*X(2, :); % Create complex value from real and imaginary parts
    f = z.^2 - z + 1; % Evaluate complex function
    F = [real(f); imag(f)]; % Separate real and imaginary parts

Then, using the initial guess "z = 1 + 2j", the function can be solved by the following command:

>> [z fval] = fsolve(@myfun, [1; 2])

which produces the results:

z =
    0.5000
    0.8661

fval =
  6.2523e-005

Therefore, the complex value that produces a local minimum of "abs(f(z))" is found to be:

z = 0.5 + i*0.8661