roots for nonlinear equation

I'm trying to solve and finding the roots for equation 2 as in the picture in such way it will give the same result if i intend to solve equation 3 ,

i tried to use fsolve command and fzero command but not reach to solution

Any suggestion commands

coding

function fval=eqns(ere)
P=(0.27);
er1=(1+1.8*10^14*1i); 
er2=(2.5+2.5*10^-3*1i); 
fval=((P).*((ere-er1)./(er1-2.*ere)))+((1-P).*((ere-er2)./(er2-2.*ere)));
end

Answer 1

[Matt J]

You need to pose the problem in terms of real numbers only,

[ere,fval]=fsolve(@eqns,[1,1]);
ere=complex(ere(1),ere(2))
fval,

function fval=eqns(ere)
    P=(0.27);

    ere=complex(ere(1),ere(2)); %<---added

    er1=(1+1.8*10^14*1i);
    er2=(2.5+2.5*10^-3*1i);
    fval=((P).*((ere-er1)./(er1-2.*ere)))+((1-P).*((ere-er2)./(er2-2.*ere)));
 
    fval=[real(fval),imag(fval)]; %<---added
end

With the initial guess [1,1], the result solves the equations quite well

ere =

   1.9685 + 0.0020i


fval =

   1.0e-08 *

   -0.1270    0.5217

[Matt J]

Ammar Ahmed's comment moved here

thanks for replying, i already got this solution but it's not what I want .

as i mentioned in the picture uploaded the Equation 2 and 3 should have the same roots

using Fsolve or Fzero Does not give up the Same roots.

Equation 2 Must have at least two roots one of them is the same root of Equation 3 the trivial equatio.

many thanks

[Catalytic]

I don't think either fsolve or you Equation 3 could ever give you numbers you can trust in double floating point arithmetic. There is not enough precision in double floats to evaluate the imaginary part of,

(3*p-1)*ere1+(2-3*p)*ere2

for the values that you've shown. The values 1.8e14 and 2.5e-3 are too far apart.

Maybe have a look at vpasolve().

Answer 2

[Walter Roberson]

There is nothing to solve for. You have specific inputs for p, er1, er2, and equation 3 has only those on the right hand side. Therefore the left hand side, ere, is completely defined as a specific numeric value.

Now take that numeric value for ere and substitute it and the p, er1, er2 into equation 2. The equation will either be satisfied or else it will not be.

Might I suggest that the real problem is to find the p value that works? If so then the solution is p = 1

With p = 0.27 then if you wanted to find one of er1 or er2 given the other, then the solution is er1 == er2