Question

c语言 复数 运算

怎么让C语言实现e的复次幂运算？

比如经典的e^(pi*i)=-1
还有e^(i)=0.5403 + 0.8415i ?
还有怎么求(1+i)^0.3这样的函数？编程需要用到这些计算

Answer 1

这个是一个列子，可以参考下
struct complex{
float rmz; //实部
float lmz;//虚部
};
//产生一个复数.
complex getAComplex(float a,float b){
complex Node=new complex();
Node.rmz=a;
Node.lmz=b;
return Node;}
//两个复数求和
complex addComplex(complex complex1,complex complex2)
{
complex Node=new complex();
Node.rmz=complex1.rmz+complex2.rmz;
Node.lmz=complex1.lmz+complex2.lmz;
return Node;
}
//求两个复数的差
complex subComplex(complex complex1,complex complex2)
{
complex Node=new complex();
Node.rmz=complex1.rmz-complex2.rmz;
Node.lmz=complex1.lmz-complex2.lmz;
return Node;
}
//求两个复数的积
complex productComplex(complex complex1,complex complex2)
{
complex Node=new complex();
Node.rmz=complex1.rmz*complex2.rmz-complex1.lmz*complex2.lmz;
Node.lmz=complex1.lmz*complex2.rmz+complex2.lmz*complex2.rmz;
return Node;
}
//求实部
float getComplexRmz(complex complex1)
{
return complex1.rmz;
}
//求虚部
float getComplexLmz(complex complex1)
{
return complex1.lmz;
}

Answer 2

是这样的：e^(ix)=cosx+isinx (x是实数)
所以，e^(i)=cos1+isin1=0.5403 + 0.8415i (1弧度=180度/3.1416=57.2956度）
(1+i)^0.3=(√2e^(iπ/4))^0.3=(2^0.15) * e^(i3π/40)
再按第一行所列公式展开