When Does The Imaginary Become Real?

The square root of 4 is 2 (or -2), and the square root of 100 is 10 (or -10). But what's the square root of -4? -100? To take the square root of negative numbers, we have to introduce the idea of √(-1) (Read: the square root of negative one). We call this i, the imaginary number.

The number i has a few interesting properties.

It cannot be combined with a real number like other real numbers. To combine i with the real number 2, you would get the result 2 + i. This is a complex number because it has both real and imaginary parts.

Its exponential powers cycle. So i to the first power is i, i to the second power (i squared) is -1, i to the third power (i cubed) is -i and i to the fourth power is 1. The pattern repeats to infinity (1-4 = 5-8 = 9-12, etc.).

Even though the imaginary number is difficult to consider in many applications, it is even possible to take its square root (this may be covered later).

One of the most interesting problems involving i that I've come across is i to the i-th power (i^i).

Two to the second power (2^2) is 4, and five to the fifth power (5^5) is 3,125, but what is i to the i-th power? Is it real, imaginary or complex? Can we even calculate it?

The answer, surprisingly, is yes, we can calculate i^i.

To do so, let's recall Euler's Formula from an earlier post.

e^(ix) = cos(x) + i sin(x)

We want to find a value for x so that we get the following:

e^(ix) = i

Since cos(x) is entirely real, we want that to equal 0.

Since sin(x) is going to be the imaginary coefficient, we want it to equal 1.

cos(x) = 0 whenever x = π/2, 3π/2, 5π/2, ...

sin(x) = 1 whenever x = π/2, 5π/2, 9π/2, ...

Any value of x = π/2, 5π/2, 9π/2, ... will work. Since π/2 is the easiest, we'll use it.

e^(iπ/2) = cos(π/2) + i sin(π/2)

e^(iπ/2) = 0 + (i)(1)

e^(iπ/2) = i

We can now rewrite i^i as [e^(iπ/2)]^i.

Recalling rules of exponents, whenever a power is raised to a power, the powers multiply.

Our two powers are iπ/2 and i. When these multiply, we get (i^2)(π/2).

As we said earlier, i^2 is -1. So our new exponent is (-π/2).

This makes our new equation:

i^i = e^(-π/2)

We no longer have an imaginary number in our exponent! We can easily calculate this and find:

e^(-π/2) = 0.207879576...

What if we'd used another value like 5π/2?

A quick calculation gives us:

e^(-5π/2) = 0.000388203204...

We got a different value! Let's try one more value. Even though -3π/2 is a negative number, it is a solution for e^(ix) = i.

The only difference is that the exponent will be positive, instead of negative, this time.

e^(3π/2) = 111.317778...

Three different values and three different answers. If we tried every value of x that solves e^(ix) = i, we'd get a different solution every time. So does this mean i^i has no solution?

In one of those peculiarities of mathematics, we've arrived at a multi-valued function. This means that i^i doesn't have one value; it has infinitely many values, and they're all positive real numbers.

In most cases, π/2 will be the value mentioned (since it is the principal value to make sin(x) = 1). 
As such, the principal value for i^i usually given will be 0.207879576...

Just remember that this is one of many, many values.

Anyway, I hope you got something out of this. If not, well, there's always next time.

If you have any questions, or if anything isn't clear, just let me know in the comments.

I came across this mathy quirk while reading the book An Imaginary Tale: The Story of √-1. (ISBN: 978-0-691-14600-3).