Exploring Complex Powers: Real Results From Imaginary Exponents

Complex numbers – made up of real and imaginary parts – often yield surprising results when raised to imaginary powers. In this article, we explore four fascinating cases:

• (-1)^i
• The i-th root of -1
• i^i
• The i-th root of i

Remarkably, all four produce real numbers – not just one, but infinitely many – thanks to the periodic properties of complex exponentiation. Let’s uncover these results using Euler’s formula and polar form.

What Are Complex Numbers

A complex number is written as:

z = a + bi

Where:

• a is the real part
• b is the imaginary part
• i is the imaginary unit, where i² = -1

Complex numbers are crucial in solving equations like x² + 1 = 0 and appear widely in physics, engineering, and advanced mathematics.

To understand complex powers, we must use Euler’s formula and represent complex numbers in polar form.

Euler’s Formula and Polar Form

Euler’s formula bridges complex exponentials and trigonometry:

e^(iθ) = cos(θ) + i sin(θ)

Any complex number can be written in polar form:

z = r e^(iθ)

Where:

• r is the magnitude (|z|)
• θ is the argument (angle with the positive real axis)

This form simplifies complex exponentiation and makes it easier to extract real results from imaginary powers.

Case 1: (-1)^i – Raising -1 to an Imaginary Power

Express -1 in polar form:

-1 = e^(iπ) (since cos(π) + i sin(π) = -1)

Now calculate:

(-1)^i = (e^(iπ))^i = e^(i²π) = e^(-π) ≈ 0.04321

This is a real number.

Due to periodicity (θ = π + 2kπ), the general form becomes:

(-1)^i = e^[-(π + 2kπ)], k ∈ ℤ

Each value of k gives a unique real number. Principal value is e^(–π).

Case 2: The i-th Root of -1

We interpret the i-th root of –1 as:

(-1)^(1/i) = (-1)^(-i)

Using polar form:

(-1)^(-i) = (e^(iπ))^(-i) = e^(-i²π) = e^π ≈ 23.1407

With periodicity (π + 2kπ), we have:

(-1)^(1/i) = e^(π+2kπ), k ∈ ℤ

Another infinite set of real numbers. Principal value is e^π.

Case 3: i^i – Imaginary Base and Exponent

The polar form of i is:

i = e^(iπ/2) (since cos(π/2) + i sin(π/2) = i)

Now compute:

i^i = (e^(iπ/2))^i = e^(i²π/2) = e^(-π/2) ≈ 0.20788 

Taking periodicity into account:

i^i = e^[-(π/2+2kπ)], k ∈ ℤ

All values are real. Principal value is e^(–π/2).

Case 4: The i-th Root of i

We rewrite the i-th root of i as:

i^(1/i) = i^(-i)

Using the polar form:

i^(-i) = (e^(iπ/2))^(-i) = e^(-i²π/2) = e^(π/2) ≈ 4.8105

Considering periodicity:

i^(1/i) = e^(π/2+2kπ), k ∈ ℤ

Again, an infinite set of real numbers. Principal value is e^(π/2).

Conclusion: Infinite Real Numbers from Imaginary Powers

Each of the four complex powers surprisingly yields real results:

• (-1)^i = e^(-π-2kπ)
• (-1)^(1/i) = e^(π+2kπ)
• i^i = e^(-π/2-2kπ)
• i^(1/i) = e^(π/2+2kπ)

These infinite families of real numbers highlight the beauty and power of Euler’s formula and complex exponentiation. What may seem like abstract or paradoxical operations actually yield precise, elegant, and real results.

It is important to note not only that all values ​​are real numbers, but also that the values ​​of the i-th root are greater than the values ​​of the i-th power of the corresponding number.