Returning the value of nth power of iota(i) using JavaScript

In mathematics, the imaginary unit i is defined as the square root of -1. When calculating powers of i, the results follow a cyclic pattern that repeats every 4 powers.

Mathematical Background

The imaginary unit i has the following properties:

i = ?(-1)
i² = -1
i³ = -i
i? = 1

Since i? = 1, the pattern repeats every 4 powers. This means we can use the modulo operator to determine the result for any power of i.

Power Pattern

JavaScript Implementation

const num = 657;

const findNthPower = (num = 1) => {
   switch(num % 4){
      case 0:
         return '1';
      case 1:
         return 'i';
      case 2:
         return '-1';
      case 3:
         return '-i';
   };
};

console.log(`i^${num} = ${findNthPower(num)}`);

i^657 = i

Testing Multiple Powers

const findNthPower = (num = 1) => {
   switch(num % 4){
      case 0:
         return '1';
      case 1:
         return 'i';
      case 2:
         return '-1';
      case 3:
         return '-i';
   };
};

// Test different powers
for(let n = 0; n <= 10; n++) {
   console.log(`i^${n} = ${findNthPower(n)}`);
}

i^0 = 1
i^1 = i
i^2 = -1
i^3 = -i
i^4 = 1
i^5 = i
i^6 = -1
i^7 = -i
i^8 = 1
i^9 = i
i^10 = -1

Alternative Implementation

We can also use an array-based approach for cleaner code:

const findNthPowerArray = (num) => {
   const powers = ['1', 'i', '-1', '-i'];
   return powers[num % 4];
};

console.log(`i^657 = ${findNthPowerArray(657)}`);
console.log(`i^1000 = ${findNthPowerArray(1000)}`);

i^657 = i
i^1000 = 1

Conclusion

Powers of the imaginary unit i follow a predictable cycle of 4. Using the modulo operator with 4, we can efficiently calculate any power of i without complex mathematical operations.