I am new to mathematica (less then a month), with the discussion available on this platform and with some tutorial available I have manage to write a code (not a smart one) to solve for the parameter values (values can be real or complex) for which the determinant is zero using Reduce function.
Objective: Search for value of 'sb' for which the det(DCoeffMat) is zero for different values of 'Pb'
(*%%% Parameters %%%%*)
ClearAll[sb, Pb, b, gama, L, w, h, Y, MI, EI, mu, rho, A, CN]
L = 10;
w = 0.02;
h = 0.02;
Y = 209*10^9;
MI = w*(h^3)/12;
EI = Y*MI;
mu = 0.001;
rho = 7800;
A = w*h;
CN = 4*Pi*mu/(Log[2*L/w] + 0.5) // N;
gama = rho*A*EI/(CN^2*L^4)
Do[B = b /. NSolve[b^4 + Pb*b^2 + sb + gama*sb^2 == 0, b] // N;
DCoeffMat = {{1, 1, 1, 1}, {B[[1]]^3 + Pb*B[[1]],
B[[2]]^3 + Pb*B[[2]], B[[3]]^3 + Pb*B[[3]],
B[[4]]^3 + Pb*B[[4]]}, {B[[1]]*Exp[B[[1]]], B[[2]]*Exp[B[[2]]],
B[[3]]*Exp[B[[3]]],
B[[4]]*Exp[B[[4]]]}, {B[[1]]^2*Exp[B[[1]]] + Pb*Exp[B[[1]]],
B[[2]]^2*Exp[B[[2]]] + Pb*Exp[B[[2]]],
B[[3]]^2*Exp[B[[3]]] + Pb*Exp[B[[3]]],
B[[4]]^2*Exp[B[[4]]] + Pb*Exp[B[[4]]]}};
DA = Det[DCoeffMat];
Print[Reduce[
DA == 0 && -1000 < Re[sb] < 1000 && -1000 < Im[sb] < 1000, sb] //
Quiet // N]
, {Pb, 0.1, 4., 0.1}]
Here is the equation used for finding B 
Here is the matrix whose determinant is equated to zero 
On running the program its showing an error.
Am I missing something???
As per the literature from where the problem is pick up it says that for pb<2.4674 the roots of sb has small real part (with non zero imaginary part) and after that its real value increases.
Thanks in advance.
NSolve
withsb
as a parameter. You're missing a;
after the// N
and beforeDCoeffMat
. And it's unlikely thatReduce
will be able to solve your equation. You will most likely need to useFindRoot
instead. $\endgroup$FindRoot
can give complex roots, e.g.,FindRoot[x^2 - 3 I, {x, 1}]
. $\endgroup$FindRoot[DA, {sb, I}]
, its working. But I want vary the initial point (in the above case I have given 0+0*I). How can I do that within the Do loop?? With the motive to find different solutions. $\endgroup$