解复数方程，哥哥姐姐们救命啊
各个哥哥姐姐，我本科毕业论文要用metlab解一个矩阵的未知数，就是矩阵中有一个未知数，求出使矩阵的行列式等于零，而且矩阵很大，不能用det做出来，即使做出来，solve函数也解不了，因为未知数是复数，矩阵行列式的值也是复数，k就是未知数，h等于零可以解，不等于零就解不出来，不知道各位有没有迭代程序。矩阵中有sin,cos。
不知道四年大学能不能毕业，早知，我就不读理科了，女孩子真吃亏！
伊玲在此先谢过了！
syms k
i=sqrt(-1);
h=0.0001;
w=6.28e6;
de1=7800;
de2=1000;
lambda1=5e10;
mu1=5e10;
mu2=0.1;
kl=w/sqrt((lambda1+2*mu1)/de1);
kt=w/sqrt(mu1/de1);
beta1=sqrt(1-kl^2/k^2);
beta2=sqrt(1-kt^2/k^2);
cy=1500;
ky=w/cy;
bian=sqrt(2*mu2/w/de2);
% m=1;
m=(1+i)/bian;
r=sqrt(ky^2-k^2);
fp=(2*k^2*mu1*beta1*de2*sin(r*h)*beta2-beta1*sin(r*h)*m*k*mu1*de2*beta2^2-k^2*mu1*de2*sin(r*h)-beta1*sin(r*h)*m^2*de2*mu2*beta2-k^2*mu1*de2*sin(r*h)*beta2^2+de2*sin(r*h)*m^2*mu2+beta1*sin(r*h)*m*k*mu1*de2)/beta2*w+(2*i*k^2*beta1^2*mu1^2*m*r*cos(r*h)-i*k^2*lambda1*mu1*m*r*cos(r*h)+4*i*k^4*mu1^2*beta1*beta2*sin(r*h)-4*i*k^2*mu1^2*beta1*beta2*m*r*cos(r*h)-i*k^4*beta1^2*lambda1*mu1*beta2^2*sin(r*h)-i*k^4*beta1^2*lambda1*mu1*sin(r*h)-2*i*k^4*beta1^2*mu1^2*sin(r*h)+2*i*k^2*beta1^2*mu1*sin(r*h)*m^2*mu2-2*i*k*mu1*r*cos(r*h)*m^2*mu2*beta2+2*i*k^2*beta1^2*mu1^2*beta2^2*m*r*cos(r*h)+i*k^4*lambda1*mu1*beta2^2*sin(r*h)-2*i*beta1*sin(r*h)*m^2*k^2*mu1*mu2*beta2-i*k*lambda1*r*cos(r*h)*m^2*mu2*beta2-i*k^2*lambda1*sin(r*h)*m^2*mu2+i*k^2*beta1^2*lambda1*sin(r*h)*m^2*mu2+i*k^2*beta1^2*lambda1*mu1*m*r*cos(r*h)-i*k^2*lambda1*mu1*beta2^2*m*r*cos(r*h)-2*i*k^4*beta1^2*mu1^2*beta2^2*sin(r*h)+2*i*k*beta1^2*mu1*r*cos(r*h)*m^2*mu2*beta2+i*k^2*beta1^2*lambda1*mu1*beta2^2*m*r*cos(r*h)+i*k*beta1^2*lambda1*r*cos(r*h)*m^2*mu2*beta2+i*k^4*lambda1*mu1*sin(r*h))/beta2
kk=solve(fp,k) 有以前做过这个算法，今年修改了一下你的程序
可是找不到根
能贴个文档把问题写的清楚些吗？看起来像是力学有关的吧？ function fp=myfun(k)
i=sqrt(-1);
h=0.0001;
w=6.28e6;
de1=7800;
de2=1000;
lambda1=5e10;
mu1=5e10;
mu2=0.1;
kl=w/sqrt((lambda1+2*mu1)/de1);
kt=w/sqrt(mu1/de1);
beta1=sqrt(1-kl^2/k^2);
beta2=sqrt(1-kt^2/k^2);
cy=1500;
ky=w/cy;
bian=sqrt(2*mu2/w/de2);
% m=1;
m=(1+i)/bian;
r=sqrt(ky^2-k^2);
fp=(2*k^2*mu1*beta1*de2*sin(r*h)*beta2-beta1*sin(r*h)*m*k*mu1*de2*beta2^2-k^2*mu1*de2*sin(r*h)-beta1*sin(r*h)*m^2*de2*mu2*beta2-k^2*mu1*de2*sin(r*h)*beta2^2+de2*sin(r*h)*m^2*mu2+beta1*sin(r*h)*m*k*mu1*de2)/beta2*w+(2*i*k^2*beta1^2*mu1^2*m*r*cos(r*h)-i*k^2*lambda1*mu1*m*r*cos(r*h)+4*i*k^4*mu1^2*beta1*beta2*sin(r*h)-4*i*k^2*mu1^2*beta1*beta2*m*r*cos(r*h)-i*k^4*beta1^2*lambda1*mu1*beta2^2*sin(r*h)-i*k^4*beta1^2*lambda1*mu1*sin(r*h)-2*i*k^4*beta1^2*mu1^2*sin(r*h)+2*i*k^2*beta1^2*mu1*sin(r*h)*m^2*mu2-2*i*k*mu1*r*cos(r*h)*m^2*mu2*beta2+2*i*k^2*beta1^2*mu1^2*beta2^2*m*r*cos(r*h)+i*k^4*lambda1*mu1*beta2^2*sin(r*h)-2*i*beta1*sin(r*h)*m^2*k^2*mu1*mu2*beta2-i*k*lambda1*r*cos(r*h)*m^2*mu2*beta2-i*k^2*lambda1*sin(r*h)*m^2*mu2+i*k^2*beta1^2*lambda1*sin(r*h)*m^2*mu2+i*k^2*beta1^2*lambda1*mu1*m*r*cos(r*h)-i*k^2*lambda1*mu1*beta2^2*m*r*cos(r*h)-2*i*k^4*beta1^2*mu1^2*beta2^2*sin(r*h)+2*i*k*beta1^2*mu1*r*cos(r*h)*m^2*mu2*beta2+i*k^2*beta1^2*lambda1*mu1*beta2^2*m*r*cos(r*h)+i*k*beta1^2*lambda1*r*cos(r*h)*m^2*mu2*beta2+i*k^4*lambda1*mu1*sin(r*h))/beta2;
>> x0=0.1;
>> options=optimset('Display','iter');
>> =fsolve(@myfun,x0,options)
不知道你的初值如何估计
随便估计x0=1+i;
计算的结果如下（不理想）
>> x0=1+1;
>> options=optimset('Display','iter');=fsolve(@myfun,x0,options)
Norm of First-order Trust-region
IterationFunc-count f(x) step optimality radius
1 2 3.34341e+080 1.67e+080 1
2 4 1.48596e+080 1 4.95e+079 1
3 6 8.35848e+079 1 2.09e+079 1
4 8 5.34941e+079 1 1.07e+079 1
5 10 3.71485e+079 1 6.19e+078 1
6 12 2.72926e+079 1 3.9e+078 1
7 14 2.08958e+079 1 2.61e+078 1
8 16 1.65101e+079 1 1.83e+078 1
9 18 1.33731e+079 1 1.34e+078 1
10 20 8.55857e+078 2.5 6.85e+077 2.5
11 22 5.94327e+078 2.5 3.96e+077 2.5
12 24 4.36633e+078 2.5 2.5e+077 2.5
13 26 3.34284e+078 2.5 1.67e+077 2.5
14 28 2.64113e+078 2.5 1.17e+077 2.5
15 30 2.13921e+078 2.5 8.56e+076 2.5
16 32 1.36888e+078 6.25 4.38e+076 6.25
17 34 9.50436e+077 6.25 2.54e+076 6.25
18 36 6.98125e+077 6.25 1.6e+076 6.25
19 38 5.34366e+077 6.25 1.07e+076 6.25
20 40 4.22093e+077 6.25 7.51e+075 6.25
21 42 3.41785e+077 6.25 5.48e+075 6.25
22 44 2.18534e+077 15.625 2.8e+075 15.6
23 46 1.51582e+077 15.625 1.62e+075 15.6
24 48 1.11212e+077 15.625 1.02e+075 15.6
25 50 8.50109e+076 15.625 6.85e+074 15.6
26 52 6.70473e+076 15.625 4.81e+074 15.6
27 54 5.4198e+076 15.625 3.51e+074 15.6
28 56 3.44776e+076 39.0625 1.79e+074 39.1
29 58 2.37653e+076 39.0625 1.04e+074 39.1
30 60 1.73061e+076 39.0625 6.54e+073 39.1
31 62 1.31138e+076 39.0625 4.38e+073 39.1
32 64 1.02395e+076 39.0625 3.08e+073 39.1
33 66 8.18348e+075 39.0625 2.24e+073 39.1
34 68 5.02781e+075 97.6563 1.15e+073 97.7
35 70 3.31323e+075 97.6563 6.65e+072 97.7
36 72 2.27888e+075 97.6563 4.19e+072 97.7
37 74 1.60689e+075 97.6563 2.81e+072 97.7
38 76 6.82038e+074 244.141 1.26e+072 244
39 78 2.264e+074 244.141 6.97e+071 244
40 80 1.01291e+074 324.785 3.96e+071 610
41 81 1.01291e+074 255.951 3.96e+071 610
42 83 6.95391e+073 63.9878 3.7e+071 64
43 85 4.30628e+073 63.9878 3.69e+071 64
44 86 4.30628e+073 63.9878 3.69e+071 64
45 88 3.16448e+073 15.9969 3.44e+071 16
46 89 3.16448e+073 39.9924 3.44e+071 40
47 91 2.69028e+073 9.99809 3.37e+071 10
48 93 2.60576e+073 9.99809 3.6e+071 10
49 95 2.22423e+073 9.99809 3.61e+071 10
50 96 2.22423e+073 9.99809 3.61e+071 10
51 98 2.04552e+073 2.49952 3.53e+071 2.5
52 99 2.04552e+073 6.24881 3.53e+071 6.25
53 101 1.99702e+073 1.5622 3.55e+071 1.56
Maximum number of function evaluations reached:
increase options.MaxFunEvals
k =
1.4548e+003 +5.3904e-001i
fval =
4.4555e+036 -3.4493e+035i
你自己再试试别的初值吧
更加详细的问题
k是一个复数，具体的数值在实部和虚部范围内都是-4000-+4000，是力学里的波导的求法 答案大约在2697.8...+0*i 附近当小数点不断减小的时候，DET可趋近于零
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