complex solution by Newton's method

#include <complex.h>
#include <float.h>
#include <math.h>
#include <stdio.h>

static inline complex double f(complex double z) {
    return z * z + 5 * z + 7;
}

static inline complex double df(complex double z) {
    return 2 * z + 5;
}

int main(void) {
    for (int i = -10; i <= 10; ++i) {
        for (int j = -10; j <= 10; ++j) {
            if (i == 0 || j == 0) {
                continue;
            }
            int c = 0;
            complex double z0 = i / 10. + j / 10. * I;
            complex double z = z0 - f(z0) / df(z0);
            for (; fabs(z - z0) > DBL_EPSILON;
                ++c, z0 = z, z = z0 - f(z0) / df(z0)) {
                ;
            }
            printf("z = %f %+f * I, c = %d, z0 = %f %+f * I\n",
                creal(z), cimag(z), c, i / 10., j / 10.);
        }
    }
}