如何在c中打印二次方程的虚根？

Question

我需要打印二次方程的虚根。 但是当我执行我的代码时，结果显示虚根是 0.00000i。 甚至我用 也一样。

大家可以帮我检查一下我加粗的代码吗？

//C program to find the root of the quadratic equation
#include<stdio.h>
#include<math.h>
#include<complex.h>

int main()
{
    double  a, b, c, x, x1, x2, disc, xr, ximg1, ximg2;
    
    printf("Please enter the value of quadratic equation, a: ");
    scanf("%lf", &a);
    printf("Please enter the value of quadratic equation, b: ");
    scanf("%lf", &b);
    printf("Please enter the value of quadratic equation, c: ");
    scanf("%lf", &c);

    if( a == 0 )
    {
        x = -(c/b);
        printf("\nThis is not a quadratic equation.\n");
        printf("x = %.3lf", x);
    }
    else{
    disc = (b*b) - 4*a*c;
        if( disc == 0 ){
             x1 = -b / 2 * a;
             x2 = -b / 2 * a;
             printf("x1 = %lf, x2 = %lf", x1, x2);
        }
        else if(disc > 0){
            x1 = ( -b + sqrt( disc ) ) / 2 * a;
            x2 = ( -b - sqrt( disc ) ) / 2 * a;
            printf("x1 = %.1lf, x2 = %.1lf", x1, x2);
        }
        else{
            ximg1 = sqrt( disc ) / 2 * a;
            ximg2 = - sqrt( disc ) / 2 * a;
            xr = - b / ( 2 * a );
            **printf("xr = %lf, ximg1 = %lfi, ximg2 = %lfi", crealf(xr), cimagf(ximg1), cimagf(ximg2));**
        }
    }

    return 0;
}

输出如下：

Please enter the value of quadratic equation, a: 4
Please enter the value of quadratic equation, b: 1
Please enter the value of quadratic equation, c: 3
xr = -0.125000, ximg1 = 0.000000i, ximg2 = 0.000000i
Process returned 0 (0x0)   execution time : 3.914 s
Press any key to continue.

Answer 1

您应该使用计算的实部和虚部将根打印为复数。不需要<complex.h> 也不需要复杂类型：

    double xr = - b / ( 2 * a );
    double ximg1 = -sqrt( -disc ) / (2 * a);
    double ximg2 = -ximg1;
    printf("x1 = %lf%+lfi, x2 = %lf%+lfi\n", xr, ximg1, xr, ximg2);

Answer 2

不需要复杂类型

当代码到达下面时，disc < 0。求否定的平方根。

        // ximg1 = sqrt( disc ) / 2 * a;
        ximg1 = sqrt( -disc ) / 2 * a;     // Use -disc
        // ximg2 = - sqrt( disc ) / 2 * a;
        ximg2 = -ximg1;                    // Simply negate ximg1
        xr = - b / ( 2 * a );
        printf("xr = %lf, ximg1 = %lfi, ximg2 = %lfi", 
        // crealf(xr), cimagf(ximg1), cimagf(ximg2));
          xr, ximg1, ximg2);

提示：使用"%e"，信息量更大。

        printf("xr = %le, ximg1 = %lei, ximg2 = %lei", 
          xr, ximg1, ximg2);

错误

而不是/ 2 * a;，在各个地方，你当然想要/ (2 * a);。

Answer 3

由于您使用了复杂的头文件，您只需要使用 csqrt() 而不是 sqrt()，

//C program to find the root of the quadratic equation
#include<stdio.h>
#include<math.h>
#include<complex.h>

int main()
{
    double  a, b, c, x, x1, x2, disc, xr, ximg1, ximg2;
    
    printf("Please enter the value of quadratic equation, a: ");
    scanf("%lf", &a);
    printf("Please enter the value of quadratic equation, b: ");
    scanf("%lf", &b);
    printf("Please enter the value of quadratic equation, c: ");
    scanf("%lf", &c);

    if( a == 0 )
    {
        x = -(c/b);
        printf("\nThis is not a quadratic equation.\n");
        printf("x = %.3lf", x);
    }
    else{
    disc = (b*b) - 4*a*c;
        if( disc == 0 ){
             x1 = -b / 2 * a;
             x2 = -b / 2 * a;
             printf("x1 = %lf, x2 = %lf", x1, x2);
        }
        else if(disc > 0){
            x1 = ( -b + csqrt( disc ) ) / 2 * a;//changed the function here
            x2 = ( -b - csqrt( disc ) ) / 2 * a;//changed the function here
            printf("x1 = %.1lf, x2 = %.1lf", x1, x2);
        }
        else{
            ximg1 = sqrt( disc ) / 2 * a;
            ximg2 = - sqrt( disc ) / 2 * a;
            xr = - b / ( 2 * a );
            **printf("xr = %lf, ximg1 = %lfi, ximg2 = %lfi", crealf(xr), cimagf(ximg1), cimagf(ximg2));**
        }
    }

    return 0;
}