复数方程（组）求解
一个复数对应2个实数。目前，Forcal中虽然没有专门求解复数方程（组）的函数，但可以借助实数算法来求解。例子1：求实数方程复数域内的全部解：x^3+2*x*x+10*x-20=0;
本例若用isolve求解，只能获得实数解：
!using["fcopt"];
f(x)=2*x^6-x^3+2*x*x+10*x-20;
isolve;
-1.543029953303134 7.105427357601002e-015
1.221035549850575 3.552713678800501e-015
用solve求解方程组，可获得复数解（需要多次求解，与实数解比较，获得复数解）：
!using["fcopt"];
c: cf(x,y)= y=2*x^6-x^3+2*x*x+10*x-20;
cc(x,y,y1,y2)= cf(x,y,&y1,&y2);
solve;
1.221035549850575 2.559801991936669e-027 2.51214793389404e-015
0.8999326566465651 1.099717348577194 3.76822190084106e-015
0.8999326566465653 -1.099717348577193 7.944109290391274e-015
-1.543029953303134 1.782340001805797e-016 1.560233745709171e-014
-0.738935454920286 -1.443073377091521 2.161031364628563e-014
-0.738935454920286 1.443073377091521 2.369955620929136e-014
本帖最后由 forcal 于 2011-1-17 20:28 编辑
例子2：求复数方程组：
(2+5i)*x1-x2^(2-3i)-exp(-x1)=0
-(x1^3)+x1*x2-exp(-x2)=0
代码：
!using["fcopt"];
c: cf(x1,x2,y1,y2)=
{
y1=(2+5i)*x1-x2^(2-3i)-exp(-x1),
y2=-(x1^3)+x1*x2-exp(-x2)
};
cc(x11,x12,x21,x22,y11,y12,y21,y22)= cf(x11,x12,x21,x22,&y11,&y12,&y21,&y22);
solve;
结果（多次求解，每组最后一个是误差）：
0.350403406122754 -0.2581172046401702 0.9031492305415151 0.2062068702236727 3.500829247998799e-014
8.343862612128539e-002 -0.1745973157301181 0.3407059687466865 -3.686653995029365 1.014802625489917e-011
-2.866006054280364e-002 -1.195892184488039e-002 1.330587481557424 -8.406178015373364 1.344455731312484e-011
0.3248221986204087 -0.8164562030925229 8.588879599246295e-002 0.2894415311777754 3.570823474245571e-011
8.663773118603429e-003 5.927370162144334e-002 2.009821712522968 -0.9744617637719372 1.073856633613092e-010
0.1575094449966286 -6.233406303049387e-003 -0.5428160523568441 -10.90400468060007 1.542485713397894e-013
0.5108186106709388 0.8884343740105763 -3.912183017503166e-002 1.868542038988761e-002 4.891644231048831e-012
-1.000140126356298 4.481460412883024e-005 -2.935625999646882e-002 4.580361872777336e-003 1.741254712865244e-012
结果可能不全，欢迎大家补充。
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