Hi Hamed,

that equation doesn't have any real root.

Best wishes.

Since the Muller method is not definite as regards the sign of the square root, it might be easier to regard Re(f(x+i*y))=0, Im(f(x+iy))=0 as a two-dimensional function of two real variables and compute the zeroes via a two-dimensional newton-raphson method. But beware: There are quite a number of zeroes of this function. This may be seen using the maple plot command (with re=Re(f), im=Im(f))

implicitplot({re(x,y) = 0,im(x,y)=0},x=0..4*Pi,y=0..4*Pi,numpoints=8000);

One is close to 0.983+i*0.822.

I agree with the comment made above by Herbert Homeier

Re( f(x+iy) ) = 0 yields 3x = sin(x).ch(y) + exp(x).cos(y)

Im( f(x+iy) ) = 0 yields 3y = cos(x).sh(y) + exp(x).sin(y)

One pair of solutions is 0,982560 (+/-) i*0,821985

yes

clc;

clear;

close all;

p0 = 2;

p1 = 2.5;

p2 = (p0+p1)/2;

TOL = 10^-15;

N0 = 100;

format long;

f = @ (x) 3*x -sin(x) -exp(x);

h1 = p1 - p0;

h2 = p2 - p1;

DELTA1 = (f(p1) - f(p0))/h1;

DELTA2 = (f(p2) - f(p1))/h2;

d = (DELTA2 - DELTA1)/(h2 + h1);

i=3;

while i N0

formatSpec = string('The method failed after N0 iterations,N0= %d \n');

fprintf(formatSpec,N0);

end