answer (Компл. числ. Нахождения функции y (x), где x Є [a, b] с шагом h. Выполнить 3 методами: for, while, do while.)

#include "stdafx.h" #include <iostream> #include <cmath> #include <complex> using namespace std; complex<double> func(int x) { complex<double > result (0,0); complex<double > tmp(x, 0); result = cos(tmp) + sqrt(tmp); return result; } int main() { //Вариант 7 нахождение функции y(x), где x Є[a, b] (-4,3)с шагом h. int a = -3, b = 2, h = 1; complex<double> result(0, 0); cout << "var1: for " <<endl; for (int x = a; x <= b; x++) { result = func(x); if (result.imag() == 0) cout << "x= " << x << " f(x)= " << result.real() << endl; else if (result.imag() == 1) cout << "x= " << x << " f(x)= " << result.real() << " + i" << endl; else cout << "x= " << x << " f(x)= " << result.real() << " + " << result.imag() << "i" << endl; } cout << endl << "var2: while " << endl; int X = a; while (X <= b) { result = func(X); if (result.imag() == 0) cout << "x= " << X << " f(x)= " << result.real() << endl; else if (result.imag() == 1) cout << "x= " << X << " f(x)= " << result.real() << " + i" << endl; else cout << "x= " << X << " f(x)= " << result.real() << " + " << result.imag() << "i" << endl; X++; } cout << endl << "var3: do while " << endl; int eks = a; do { result = func(eks); if (result.imag() == 0) cout << "x= " << eks << " f(x)= " << result.real() << endl; else if (result.imag() == 1) cout << "x= " << eks << " f(x)= " << result.real() << " + i" << endl; else cout << "x= " << eks << " f(x)= " << result.real() << " + " << result.imag() << "i" << endl; eks++; } while (eks <= b); return 0; }