因为有matlab代码需要在opencv上实现，遇到了./的问题，在网上参考了许多代码，但是与matlab的结果不符，于是自己着手实现了一下，与matlab验证证明是正确的。
Mat cmplxDivision(Mat m1, Mat m2) {
int rows = m2.rows;//矩阵的行数以及列数int cols = m2.cols;Mat m1_2[2], m2_2[2];//用于存放矩阵的实部和虚部Mat m1_rea, m1_ima, m2_rea, m2_ima;//用于分开存放实部和虚部，便于计算split(m1, m1_2);split(m2, m2_2);m1_rea = m1_2[0];m1_ima = m1_2[1];m2_rea = m2_2[0];m2_ima = m2_2[1];Mat result1(rows, cols, CV_64FC1);//存放用于计算的实部Mat result2(rows, cols, CV_64FC1);//存放用于计算的虚部//(a+bi)/(c+di)=(ac+bd)/(c^2+d^2) +((bc-ad)/(c^2+d^2))ifor (int i = 0; i < rows; i++){for (int j = 0; j < cols; j++){result1.at<double>(i, j) = (m1_rea.at<double>(i, j)*m2_rea.at<double>(i, j) + m1_ima.at<double>(i, j)*m2_ima.at<double>(i, j)) / (m2_rea.at<double>(i, j)*m2_rea.at<double>(i, j)+ m2_ima.at<double>(i, j)*m2_ima.at<double>(i, j));result2.at<double>(i, j) = (m1_ima.at<double>(i, j)*m2_rea.at<double>(i, j) - m1_rea.at<double>(i, j)*m2_ima.at<double>(i, j)) / (m2_rea.at<double>(i, j)*m2_rea.at<double>(i, j) + m2_ima.at<double>(i, j)*m2_ima.at<double>(i, j));}}Mat result_2[2] = { result1 ,result2 };Mat result;merge(result_2, 2, result);return result;}