Complex Number Multiplication - Practice Coding Problems

Complex Number Multiplication - Problem

Complex numbers are mathematical objects that extend real numbers by introducing the imaginary unit i, where i² = -1. They're widely used in engineering, physics, and computer graphics for representing rotations and oscillations.

In this problem, you're given two complex numbers as strings in the format "a+bi" or "a-bi", where:

a is the real part (integer between -100 and 100)
b is the imaginary part (integer between -100 and 100)
i represents the imaginary unit

Your task is to multiply these two complex numbers and return the result as a string in the same format.

Mathematical Formula: (a + bi) × (c + di) = (ac - bd) + (ad + bc)i

For example: "1+1i" × "1+1i" = "0+2i" because (1×1 - 1×1) + (1×1 + 1×1)i = 0 + 2i

Input & Output

example_1.py — Basic Multiplication

$ Input: num1 = "1+1i", num2 = "1+1i"

› Output: "0+2i"

💡 Note: Using formula (a+bi)×(c+di) = (ac-bd)+(ad+bc)i: (1+1i)×(1+1i) = (1×1-1×1)+(1×1+1×1)i = 0+2i

example_2.py — Negative Result

$ Input: num1 = "1+-1i", num2 = "1+-1i"

› Output: "0+-2i"

💡 Note: With negative imaginary parts: (1-1i)×(1-1i) = (1×1-(-1)×(-1))+(1×(-1)+(-1)×1)i = (1-1)+(-1-1)i = 0-2i

example_3.py — Pure Real Numbers

$ Input: num1 = "2+0i", num2 = "3+0i"

› Output: "6+0i"

💡 Note: When imaginary parts are zero, it behaves like regular multiplication: (2+0i)×(3+0i) = (2×3-0×0)+(2×0+0×3)i = 6+0i

Visualization

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Understanding the Visualization

Input Representation

Plot both complex numbers as points on the complex plane (real axis = x, imaginary axis = y)

Component Extraction

Identify the real and imaginary parts of each complex number

Formula Application

Apply the multiplication formula: real_result = ac-bd, imag_result = ad+bc

Result Plotting

Plot the resulting complex number and format as string output

Key Takeaway

🎯 Key Insight: Complex number multiplication is essentially a combination of algebraic expansion and the fundamental property that i² = -1, which transforms what could be a complex geometric operation into straightforward arithmetic.

Time & Space Complexity

Time Complexity

⏱️

O(n)

Where n is the length of the input strings. We need to parse each character once to extract components.

✓ Linear Growth

Space Complexity

O(1)

Only using a constant amount of extra space to store the parsed components and result.

✓ Linear Space

Constraints

The real and imaginary parts are integers in the range [-100, 100]
The input strings will always be in the format "a+bi" or "a-bi"
The imaginary part will always end with the character 'i'
Both input strings represent valid complex numbers

Asked in

G Google 25 ⊞ Microsoft 18 a Amazon 15 f Meta 12

The optimal approach is direct mathematical calculation using the complex multiplication formula. Parse both input strings to extract real and imaginary components, apply the formula (a+bi)×(c+di) = (ac-bd)+(ad+bc)i, and format the result. This achieves O(n) time complexity where n is the string length, with O(1) space complexity.

Common Approaches

Approach	Time	Space	Notes
✓ Direct String Parsing and Mathematical Calculation	O(n)	O(1)	Parse both complex number strings, extract components, apply multiplication formula, format result

Direct String Parsing and Mathematical Calculation — Algorithm Steps

Parse the first complex number string to extract real part a and imaginary part b
Parse the second complex number string to extract real part c and imaginary part d
Calculate the real part of result: ac - bd
Calculate the imaginary part of result: ad + bc
Format the result as a string in the required format

Visualization

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Step-by-Step Walkthrough

Parse Input Strings

Extract real and imaginary components from both input strings

Apply Formula

Use (a+bi)×(c+di) = (ac-bd)+(ad+bc)i to calculate result

Format Output

Convert result back to required string format

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char* complexNumberMultiply(char* num1, char* num2) {
    // Parse first complex number
    char* lastPlus1 = strrchr(num1, '+');
    char* lastMinus1 = strrchr(num1 + 1, '-');
    char* split1 = (lastPlus1 > lastMinus1) ? lastPlus1 : lastMinus1;
    
    char realStr1[10], imagStr1[10];
    strncpy(realStr1, num1, split1 - num1);
    realStr1[split1 - num1] = '\0';
    strcpy(imagStr1, split1);
    imagStr1[strlen(imagStr1) - 1] = '\0'; // Remove 'i'
    
    int real1 = atoi(realStr1);
    int imag1 = atoi(imagStr1);
    
    // Parse second complex number
    char* lastPlus2 = strrchr(num2, '+');
    char* lastMinus2 = strrchr(num2 + 1, '-');
    char* split2 = (lastPlus2 > lastMinus2) ? lastPlus2 : lastMinus2;
    
    char realStr2[10], imagStr2[10];
    strncpy(realStr2, num2, split2 - num2);
    realStr2[split2 - num2] = '\0';
    strcpy(imagStr2, split2);
    imagStr2[strlen(imagStr2) - 1] = '\0';
    
    int real2 = atoi(realStr2);
    int imag2 = atoi(imagStr2);
    
    // Apply multiplication formula
    int resultReal = real1 * real2 - imag1 * imag2;
    int resultImag = real1 * imag2 + imag1 * real2;
    
    // Format result
    char* result = (char*)malloc(50);
    sprintf(result, "%d%s%di", resultReal, (resultImag >= 0 ? "+" : ""), resultImag);
    return result;
}

int main() {
    char num1[20], num2[20];
    scanf("%s %s", num1, num2);
    printf("%s\n", complexNumberMultiply(num1, num2));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Where n is the length of the input strings. We need to parse each character once to extract components.

✓ Linear Growth

Space Complexity

O(1)

Only using a constant amount of extra space to store the parsed components and result.

✓ Linear Space

Constraints

The real and imaginary parts are integers in the range [-100, 100]
The input strings will always be in the format "a+bi" or "a-bi"
The imaginary part will always end with the character 'i'
Both input strings represent valid complex numbers

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Direct String Parsing and Mathematical Calculation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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