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Formatted question description: https://leetcode.ca/all/537.html

# 537. Complex Number Multiplication (Medium)

Given two strings representing two complex numbers.

You need to return a string representing their multiplication. Note i2 = -1 according to the definition.

Example 1:

Input: "1+1i", "1+1i"
Output: "0+2i"
Explanation: (1 + i) * (1 + i) = 1 + i2 + 2 * i = 2i, and you need convert it to the form of 0+2i.


Example 2:

Input: "1+-1i", "1+-1i"
Output: "0+-2i"
Explanation: (1 - i) * (1 - i) = 1 + i2 - 2 * i = -2i, and you need convert it to the form of 0+-2i.


Note:

1. The input strings will not have extra blank.
2. The input strings will be given in the form of a+bi, where the integer a and b will both belong to the range of [-100, 100]. And the output should be also in this form.

## Solution 1.

• class Solution {
public String complexNumberMultiply(String a, String b) {
int aReal = Integer.parseInt(a.substring(0, a.indexOf('+')));
int aImag = Integer.parseInt(a.substring(a.indexOf('+') + 1, a.indexOf('i')));
int bReal = Integer.parseInt(b.substring(0, b.indexOf('+')));
int bImag = Integer.parseInt(b.substring(b.indexOf('+') + 1, b.indexOf('i')));
int mulReal = aReal * bReal - aImag * bImag;
int mulImag = aReal * bImag + bReal * aImag;
return mulReal + "+" + mulImag + "i";
}
}

############

class Solution {
public String complexNumberMultiply(String num1, String num2) {
String[] c1 = num1.split("\\+|i");
String[] c2 = num2.split("\\+|i");
int a = Integer.parseInt(c1[0]);
int b = Integer.parseInt(c1[1]);
int c = Integer.parseInt(c2[0]);
int d = Integer.parseInt(c2[1]);
return String.format("%d+%di", a * c - b * d, a * d + c * b);
}
}

• // OJ: https://leetcode.com/problems/complex-number-multiplication/
// Time: O(N)
// Space: O(N)
class Solution {
private:
vector<int> parse(string &s) {
vector<int> ans(2, 0);
ans[0] = stoi(s);
int i = s.find_first_of("+");
ans[1] = stoi(s.substr(i + 1));
return ans;
}
public:
string complexNumberMultiply(string a, string b) {
auto m = parse(a), n = parse(b);
int x = m[0] * n[0] - m[1] * n[1], y = m[1] * n[0] + m[0] * n[1];
return to_string(x) + "+" + to_string(y) + "i";
}
};

• class Solution:
def complexNumberMultiply(self, num1: str, num2: str) -> str:
a, b = map(int, num1[:-1].split('+'))
c, d = map(int, num2[:-1].split('+'))
return f'{a * c - b * d}+{a * d + c * b}i'

############

class Solution(object):
def complexNumberMultiply(self, a, b):
"""
:type a: str
:type b: str
:rtype: str
"""
(ar, ac), (br, bc) = map(int, a[:-1].split("+")), map(int, b[:-1].split("+"))
return "{}+{}i".format(str(ar * br - ac * bc), str(ar * bc + br * ac))


• func complexNumberMultiply(num1, num2 string) string {
parse := func(num string) (a, b int) {
i := strings.IndexByte(num, '+')
a, _ = strconv.Atoi(num[:i])
b, _ = strconv.Atoi(num[i+1 : len(num)-1])
return
}
a, b := parse(num1)
c, d := parse(num2)
return fmt.Sprintf("%d+%di", a*c-b*d, a*d+b*c)
}

• function complexNumberMultiply(num1: string, num2: string): string {
let arr1 = num1.split('+'),
arr2 = num2.split('+');
let r1 = Number(arr1[0]),
r2 = Number(arr2[0]);
let v1 = Number(arr1[1].substring(0, arr1[1].length - 1)),
v2 = Number(arr2[1].substring(0, arr2[1].length - 1));
let ansR = r1 * r2 - v1 * v2;
let ansV = r1 * v2 + r2 * v1;
return ${ansR}+${ansV}i;
}