本帖最后由 zhaocb 于 2015-7-18 16:22 编辑

各位大神，我需要解一个超越方程，'0=(r^2+(50*m*(2*pi*k*C)^2))*tanh(r*0.04)+i*r*(50-m)*2*pi*k*C'其中r为未知量。但结果只有0，但实际情况是有复数根，而且复数根才是我要的结果；我也试了fsolve解，但初始值不知道是多少，也得不到正确的答案。我也怀疑方程本身没有复数解，但是有大牛用其他代码可以得到正确的复数解，但我有看不懂那是什么语言。。。求助

Matlab

clear;

clc;

S0=load ('C:\Users\Zhang\Desktop\20150716z\S40.txt');

emp=load ('C:\Users\Zhang\Desktop\20150716z\emp.txt');

sfilm=0.23025851.*S0(:,2)+i*pi.*(1-S0(:,3)/180);

semp=0.23025851.*emp(:,2)+i*pi.*(1-emp(:,3)/180);

w=2*pi.*emp(:,1);

sp=sfilm./semp;

A=50*(1+sp)./(1-sp);

l0=0.04;

ZL=50;

for a=1:801;

m=A(a);

k=w(a);

C=5.8155E-11;

Lext=0.74332929;

syms r

equ=sprintf('0=(r^2+(50*m*(2*pi*k*C)^2))*tanh(r*0.04)+i*r*(50-m)*2*pi*k*C');

equ=subs(equ);

r=subs(solve(equ,r))；

end

save C:\Users\Zhang\Desktop\20150716z\111.txt ZZ -ascii;

不明代码(求大神告知这是什么语言)

****** Mathematica script to solve eq (4)********

Clear[fun,f,zexp,g,gout,z,Cap,g0,goini,nf,nc,k,m];

zexp=Import["zexp_r.dat","Table"]+\[ImaginaryI]Import["zexp_i.dat","Table"];

f=Import["frec.dat","Table"][[All,1]];

nf=Dimensions[f][[1]];

nc=Dimensions[zexp][[2]];

\!\(\(Cap = 4.6366*10\^\(-11\);\)\)

l=0.015;

g0ini=10+15\[ImaginaryI];

\!\(\(fun[g_, f_, z_]:= \((g*g + z*50*\((Cap*2*Pi*f)\)\^2)\)*

Tanh[g*l] + \ \[ImaginaryI]\ \((50 -z)\)*Cap*2*Pi*f*g;\)\)

gout=Table[0,{k,1,nf},{m,1,nc}];

Do[g0=g0ini;

Do[gout[[k,m]]=g/.FindRoot[fun[g,f[[k]],zexp[[k,m]]]==0,{g,g0}];

g0=gout[[k,m]],{k,1,nf}],{m,1,nc}]

stmp=OpenWrite["g_real.dat",FormatType\[Rule]TableForm];

SetOptions["g_real.dat",PageWidth\[Rule]Infinity];

Write[stmp,TableForm[Re[gout],TableSpacing->{0,1}]];

Close[stmp];

stmp=OpenWrite["g_imag.dat",FormatType\[Rule]TableForm];

SetOptions["g_imag.dat",PageWidth\[Rule]Infinity];

Write[stmp,TableForm[Im[gout],TableSpacing->{0,1}]];

Close[stmp];