Raising an Imaginary Number to a Power

Raising an imaginary number to a power involves multiplying the power of its coefficient by the power of the imaginary unit: $$ (ai)^n = a^n \cdot i^n $$ The power of the imaginary unit follows a repeating cycle of four: $$ i^0 = 1 \\ i^1 = i \\ i^2 = -1 \\ i^3 = -i $$ The cycle resets every four terms.

The nth power of the imaginary unit can always be reduced to an exponent between 0 and 3.

$$ i^0 = 1 \\ i^1 = i \\ i^2 = -1 \\ i^3 = -i $$

$$ i^4 = 1 \\ i^5 = i \\ i^6 = -1 \\ i^7 = -i $$

$$ i^8 = 1 \\ i^9 = i \\ i^{10} = -1 \\ i^{11} = -i $$

From this, we can conclude:

Powers with even exponents result in a real number.
Powers with odd exponents result in an imaginary number.

How to Compute the nth Power

For exponents $ n > 3 $, the power of the imaginary unit is equivalent to the power with an exponent equal to the remainder of dividing $ n $ by 4.

$$ i^n = i^{4k} \cdot i^r $$

Since any power of $ i $ with an exponent that’s a multiple of 4 is always equal to 1:

$$ i^n = 1 \cdot i^r $$

$$ i^n = i^r $$

Here, $ r $ is the remainder when $ n $ is divided by 4.

Example: The imaginary unit raised to the power of $ 14 $, $ i^{14} $, simplifies to $ i^2 $ because $ 14 \div 4 = 3 $ with a remainder of $ r = 2 $. Specifically: $$ i^{14} = i^{4 \cdot 3 + 2} $$ Using the properties of exponents: $$ i^{14} = (i^4)^3 \cdot i^2 $$ Since $ i^4 = 1 $, this becomes: $$ i^{14} = 1^3 \cdot i^2 $$ And since $ 1^3 = 1 $: $$ i^{14} = i^2 $$ Finally, recalling that $ i^2 = -1 $: $$ i^{14} = -1 $$ Thus, raising $ i $ to any power reduces to finding the equivalent exponent between $ 0 $ and $ 3 $, thanks to the cyclic nature of the powers of $ i $.

A Practical Example
The Proof

A Practical Example

Let’s compute the square of the imaginary number $ 5i $:

$$ z^2=(5i)^2 $$

Using the exponent rule $ (a \cdot b)^n = a^n \cdot b^n $:

$$ z^2=5^2 \cdot i^2 $$

$$ z^2=25 \cdot i^2 $$

Knowing that $ i^2 = -1 $:

$$ z^2=25 \cdot (-1) $$

$$ z^2=-25 $$

Example 2

Let’s calculate the cube of $ 2i $:

$$ z^3 = (2i)^3 $$

$$ z^3 = 2^3 \cdot i^3 $$

$$ z^3 = 8 \cdot i^3 $$

$$ z^3 = 8 \cdot i^2 \cdot i $$

$$ z^3 = 8 \cdot (-1) \cdot i $$

$$ z^3 = -8i $$

Example 3

Now, let’s compute $ 5i $ raised to the 14th power:

$$ z^{14} = (5i)^{14} $$

$$ z^{14} = 5^{14} \cdot i^{14} $$

The power of $ i $ simplifies to the remainder of $ 14 \div 4 $, which is 2, so $ i^{14} = i^2 $.

$$ z^{14} = 5^{14} \cdot i^{2} $$

$$ z^{14} = 5^{14} \cdot (-1) $$

$$ z^{14} = -5^{14} $$

Note: Every power of an imaginary number can be reduced to an equivalent power with an exponent between 0 and 3. For instance, $ i^{14} $ simplifies to $ i^{12} \cdot i^2 $: $$ i^{14} = i^{12} \cdot i^2 = (i^4)^3 \cdot i^2 = 1^3 \cdot i^2 = i^2 $$.

The Proof

Consider the imaginary unit:

$$ i = (0,1) $$

Any number raised to the power of 0 equals 1:

$$ i^0 = (0,1)^0 = 1 $$

Any number raised to the power of 1 is itself:

$$ i^1 = (0,1)^1 = (0,1) = i $$

The square of the imaginary unit equals -1.

$$ i^2 = (0,1)^2 = -1 $$

Proof: To calculate the square, apply the complex number multiplication rule: $$ (0,1)^2 = (0,1) \cdot (0,1) = (0 \cdot 0 - 1 \cdot 1, 0 \cdot 1 + 1 \cdot 0) = (0 - 1, 0 + 0) = (-1,0) = -1 $$ The result is the real number -1.

The cube of the imaginary unit equals -i.

$$ i^3 = (0,1)^3 $$

Using the exponent rule $ i^3 = i^2 \cdot i $:

$$ i^3 = i^2 \cdot i $$

Since $ i^2 = -1 $:

$$ i^3 = (-1) \cdot i $$

$$ i^3 = -i $$

With an exponent of 4, the cycle resets.

$$ i^4 = i^2 \cdot i^2 $$

$$ i^4 = (-1) \cdot (-1) $$

$$ i^4 = 1 $$

Note: Imaginary numbers raised to even exponents produce real numbers, while those raised to odd exponents yield imaginary numbers.

And the pattern continues.