本题要求编写程序，计算2个复数的和、差、积、商。
输入格式：
输入在一行中按照“a1 b1 a2 b2”的格式给出2个复数C1=a1+b1*i和C2=a2+b2*i的实部和虚部。题目保证C2不为0。
输出格式：
分别在4行中按照“(a1+b1i) 运算符 (a2+b2i) = 结果”的格式顺序输出2个复数的和、差、积、商，数字精确到小数点后1位。如果结果的实部或者虚部为0，则不输出。如果结果为0，则输出0.0。
输入样例1：2 3.08 -2.04 5.06输出样例1：
(2.0+3.1i) + (-2.0+5.1i) = 8.1i (2.0+3.1i) - (-2.0+5.1i) = 4.0-2.0i (2.0+3.1i) * (-2.0+5.1i) = -19.7+3.8i (2.0+3.1i) / (-2.0+5.1i) = 0.4-0.6i输入样例2：
1 1 -1 -1.01输出样例2：
(1.0+1.0i) + (-1.0-1.0i) = 0.0 (1.0+1.0i) - (-1.0-1.0i) = 2.0+2.0i (1.0+1.0i) * (-1.0-1.0i) = -2.0i (1.0+1.0i) / (-1.0-1.0i) = -1.0
解：
import java.util.Scanner;
public class Main{
/**
*（方法说明）
*@param 无
*@return 无
*@throws 无
*/
public static void main(String[] args){
Scanner scanner = new Scanner(System.in);
//输入在一行中按照“a1 b1 a2 b2”的格式给出2个复数C1=a1+b1*i和C2=a2+b2*i的实部和虚部
double a1 = scanner.nextDouble();
double b1 = scanner.nextDouble();
double a2 = scanner.nextDouble();
double b2 = scanner.nextDouble();
//获取b1 ,b2的正负号
String signB1="+";
String signB2="+";
if(b1<0)
signB1="";
if(b2<0)
signB2="";
double addA = a1+a2;
double addB = b1+b2;
double subtractionA = a1-a2;
double subtractionB = b1-b2;
double multiplicationA =a1*a2-b1*b2;
double multiplicationB =a1*b2+a2*b1;
double divisionA =( a1*a2+b1*b2 )/(Math.pow(a2, 2)+Math.pow(b2, 2));
double divisionB =( b1*a2-a1*b2)/(Math.pow(a2, 2)+Math.pow(b2, 2));
if(Math.abs(addA)>=0.05 && Math.abs(addB)>=0.05)
if(addB>0)
System.out.printf("(%.1f"+signB1+"%.1fi) + (%.1f"+signB2+"%.1fi) = %.1f+%.1fi",a1,b1,a2,b2,addA,addB);
else
System.out.printf("(%.1f"+signB1+"%.1fi) + (%.1f"+signB2+"%.1fi) = %.1f%.1fi",a1,b1,a2,b2,addA,addB);
else if(Math.abs(addA)<0.05 && Math.abs(addB)<0.05)
System.out.printf("(%.1f"+signB1+"%.1fi) + (%.1f"+signB2+"%.1fi) = 0.0",a1,b1,a2,b2);
else if(Math.abs(addA)<0.05)
System.out.printf("(%.1f"+signB1+"%.1fi) + (%.1f"+signB2+"%.1fi) = %.1fi",a1,b1,a2,b2,addB);
else
System.out.printf("(%.1f"+signB1+"%.1fi) + (%.1f"+signB2+"%.1fi) = %.1f",a1,b1,a2,b2,addA);
System.out.println("");
if(Math.abs(subtractionA)>=0.05 && Math.abs(subtractionB)>=0.05)
if(subtractionB>0)
System.out.printf("(%.1f"+signB1+"%.1fi) - (%.1f"+signB2+"%.1fi) = %.1f+%.1fi",a1,b1,a2,b2,subtractionA,subtractionB);
else
System.out.printf("(%.1f"+signB1+"%.1fi) - (%.1f"+signB2+"%.1fi) = %.1f%.1fi",a1,b1,a2,b2,subtractionA,subtractionB);
else if(Math.abs(subtractionA)<0.05 && Math.abs(subtractionB)<0.05)
System.out.printf("(%.1f"+signB1+"%.1fi) - (%.1f"+signB2+"%.1fi) = 0.0",a1,b1,a2,b2);
else if(Math.abs(subtractionA)<0.05)
System.out.printf("(%.1f"+signB1+"%.1fi) - (%.1f"+signB2+"%.1fi) = %.1fi",a1,b1,a2,b2,subtractionB);
else
System.out.printf("(%.1f"+signB1+"%.1fi) - (%.1f"+signB2+"%.1fi) = %.1f",a1,b1,a2,b2,subtractionA);
System.out.println("");
if(Math.abs(multiplicationA)>=0.05 && Math.abs(multiplicationB)>=0.05)
if(multiplicationB>0)
System.out.printf("(%.1f"+signB1+"%.1fi) * (%.1f"+signB2+"%.1fi) = %.1f+%.1fi",a1,b1,a2,b2,multiplicationA ,multiplicationB);
else
System.out.printf("(%.1f"+signB1+"%.1fi) * (%.1f"+signB2+"%.1fi) = %.1f%.1fi",a1,b1,a2,b2,multiplicationA ,multiplicationB);
else if(Math.abs(multiplicationA )<0.05 && Math.abs(multiplicationB)<0.05)
System.out.printf("(%.1f"+signB1+"%.1fi) * (%.1f"+signB2+"%.1fi) = 0.0",a1,b1,a2,b2);
else if(Math.abs(multiplicationA )<0.05)
System.out.printf("(%.1f"+signB1+"%.1fi) * (%.1f"+signB2+"%.1fi) = %.1fi",a1,b1,a2,b2,multiplicationB);
else
System.out.printf("(%.1f"+signB1+"%.1fi) * (%.1f"+signB2+"%.1fi) = %.1f",a1,b1,a2,b2,multiplicationA );
System.out.println("");
if(Math.abs(divisionA)>=0.05 && Math.abs(divisionB)>=0.05)
if(divisionB>0)
System.out.printf("(%.1f"+signB1+"%.1fi) / (%.1f"+signB2+"%.1fi) = %.1f+%.1fi",a1,b1,a2,b2,divisionA ,divisionB);
else
System.out.printf("(%.1f"+signB1+"%.1fi) / (%.1f"+signB2+"%.1fi) = %.1f%.1fi",a1,b1,a2,b2,divisionA ,divisionB);
else if(Math.abs(divisionA )<0.05 && Math.abs(divisionB)<0.05)
System.out.printf("(%.1f"+signB1+"%.1fi) / (%.1f"+signB2+"%.1fi) = 0.0",a1,b1,a2,b2);
else if(Math.abs(divisionA )<0.05)
System.out.printf("(%.1f"+signB1+"%.1fi) / (%.1f"+signB2+"%.1fi) = %.1fi",a1,b1,a2,b2,divisionB);
else
System.out.printf("(%.1f"+signB1+"%.1fi) / (%.1f"+signB2+"%.1fi) = %.1f",a1,b1,a2,b2,divisionA );
}
}
原文：http://blog.csdn.net/ch717828/article/details/44599817