COMPLEX  ROOTS
Consider the equation f(z) = 0 with a complex independent variable z = x + iy.
The equation f(z) = 0 is also can be written as sum of real and imaginary parts in the form  
f(z) = u(x,y) + i v(x,y) = 0
Now finding the root of f(z) = 0  is nothing but finding the roots of two simultaneous equations
u(x,y) = 0
v(x,y) = 0
Let (x0, y0) is an initial approximation to the solution of the above system and (x0+ Dx, y0+Dy) be the exact solution. Then we have

u(x0+ Dx, y0+Dy) = 0 and v(x 0+Dx, y0+ Dy) = 0.
Now by Taylor series expansion about (x 0, y0) we have
u(x0, y0) + Dxux(x0, y0) + Dyuy(x0, y0) + . . . = 0
v(x0, y0) + Dxvx(x0, y0) + Dyvy(x0, y0) + . . . = 0
If we neglect the second and higher degree terms in Dx and D y and then solving the above system gives

  uvy - vuy   
 uxv - uvx   
Dx =  -
   Dy =  - 


      J

      J
at (x0, y0) and the Jacobian J = uxvy - uyvx .
Thus an improved approximation to the exact solution can be written as 
x1 = x0+ Dx,  y1 = y0+D y
Or ingeneral the iterative scheme can be written as  
xi+1 = xi + Dxi  and yi+1 = yi + Dyi   for  i = 0, 1, 2, . . .
Alternatively one can also use

         z i+1 = zi -  f(zi)                       for i = 0, 1, 2, . . .
f '(zi)
This is equivalent to the scheme defined above if f(z) is an analytic function of z.
The only difference is the initial approximation z0 must be a complex number and complex arithmetic has to be used throughout.
Example - 1 :
Find a root of z3 + 1 = 0.
We have f(z) = z3 + 1 = (x + iy)3 + 1 = 0


                        x3 + 3x2yi - 3xy2 + (iy)3 +1 = 0                          (... z = x + iy)

                       (x3 - 3xy2 +1) + (3x2y - y3)i = 0

=>                   u  =  x3 - 3xy2 +1  =  0          v  =  3x2y - y3  = 0

                        ux = 3x2 -3y2,       uy = -6xy,       vx = 6xy,         vy = 3x 2 -3y2

                        J = uxvy - uyvx = (3x2 - 3y2)2 + 36x2y2
                                                = (3x2 + 3y2)2 = 9(x2 + y 2)2
 

Now the  iterative scheme with the initial approximation (x0, y0) = (0, 0) is
xi+1 =  xi -    uvy - vuy  
      J

 
yi+1 =  yi -   uxv - uvx      i = 0, 1, 2, . . .
     J

 
   xi+1 =  xi -   (xi3 - 3xiyi 2 +1)(3xi2 - 3yi2) + 6x iyi(3x2y - y3)  
                           9(x2 + y2)2

 
   yi+1 =  yi -   (3x2y - y3)(3xi2 - 3yi2) - (xi3 - 3xi yi2 +1)6xiyi                 i = 0, 1, 2, . . .
                         9(x2 + y2)2

 
i
0
1
2
3
4
5
 6
 7
xi
0.25
0.16667 
0.1522 
0.1926 
0.3193 
 0.4925
0.4998 
 0.4999
yi
0.25
 2.83333
1.8937
1.2772
 0.9104
 0.8306
 0.8674
 0.8660
So one of the roots of  z3 + 1 = 0  is 0.499 + 0.8660i.
Solution of Transcendental Equations | Solution of Linear System of Algebraic Equations | Interpolation & Curve Fitting
Numerical Differentiation & Integration | Numerical Solution of Ordinary Differential Equations
Numerical Solution of Partial Differential Equations