How To Really Solve 1ˣ = -1?

‘1ˣ = -1’ is a weird-looking equation. What possible good does solving this equation get us? Besides, didn’t we all learn that 1 raised to any power always equals 1?

1⁰ = 1 (refer to this article for the reasoning behind this)

1⁵⁹⁹⁹ = 1

1^∞ = 1

So, what gives?

You see, I recently wrote a couple of essays on how to intuitively understand Euler’s identity and how to calculate the square root of i. Following this, reader Dan Foley suggested that I demonstrate how Euler’s general formula could be used to compute nth roots of unity (among other cool applications).

Well, that’s where the equation ‘1ˣ = -1’ comes in. Solving this equation serves as an intuitive precursor to Euler’s formula and its applications.

In this essay, I will be solving ‘1ˣ = -1’ by splitting it into two parts of a puzzle: the left-hand side (LHS) and the right-hand side (RHS). Let’s get started.

This essay is supported by Generatebg

The Left-Hand Side of 1ˣ = -1

The left-hand side of this equation is just 1ˣ. Let us forget the ‘x’ for now, and focus just on the ‘1’. In my previous essay, I showed how any real or imaginary number can be represented as a complex number.

Such a complex number has a real part and an imaginary part as shown below. Here, x and y are cartesian coordinates along the complex plane, whereas r and ψ are the corresponding polar coordinates.

In the same essay, I showed how such a complex number can be visualized as a vector on a complex plane. Let us now consider a vector from 0 to 1 along the positive real axis. Initially, its angle encloses zero radians (to the positive real axis in the anticlockwise direction) and its magnitude is equal to 1.

Vector Rotation

Now, suppose we rotate this vector by 2π radians (or 360°) around a circle. By doing this, we end up where we started, and the complex number in polar coordinates yields the same value as it did for 0 radians. In fact, the value remains the same for any angle ψ that is an even multiple of π.

As you can see, we can generalize the statement as 1 = e^(i*2*k*π), where k can take any non-zero integer value (you’ll see why we exclude zero later). Now that we have come this far, let us remember to bring in the ‘x’ we chose to ignore in the beginning.

This concludes the left-hand side of our puzzle. Let us now proceed to the right-side.

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The Right-Hand Side of 1ˣ = -1

On the right-hand side of the equation, we just have -1. To work with this, let us consider the same vector as before from 0 to 1 on the complex plane. 

If we rotate this vector by π radians (or 180°), we arrive at a complex number that yields a value of -1. It is important to note the equation we obtain here. It is none other than the famous Euler’s identity.

Before we get sidetracked by the beauty of Euler’s identity, let us continue onward. If we rotate the original vector by any angle ψ that is a non-zero odd integer multiple of π, we end up getting -1 every time.

We can generalize this result as: 

e^{(2n+1)*i* π)} = -1.

This concludes the right-hand side of our puzzle. Let us now proceed with solving our main equation.

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The Solution to 1ˣ = -1

Since we have already worked out the LHS and RHS of the equation beforehand, we could just plug in the complex forms of 1 and -1 straightaway.

Taking natural logarithm on both sides, we obtain the following equation:

Now, it becomes clear as to why we chose to exclude zero for k (for details, refer to this article). Besides, we already know that e⁰ = 1. So, it is explicitly clear that k = 0 is not a candidate solution for the original equation.

When we consider the lowest non-negative integer values for n and k respectively, we arrive at n = 0 and k = 1 respectively. By doing this, we are just restricting ourselves to minimal rotation in the anticlockwise direction in this case. Consequently, we could solve the above equation as follows.

At first glance, this result looks strange since it is not an integer. However, look at what happens when we plug this value of x into the LHS of the main equation’s complex form:

We obtain the RHS of the equation, and what’s more, the Euler’s identity emerges rather elegantly!

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Mathematical Verification

To be fair, x = ½ is not a unique solution. Nonetheless, let’s plug it into the original equation, and see if our process helped us get a valid result:

There you go. Hence, the result that we got makes sense.

Final Remark

In the end, we just used the equation ‘1ˣ = -1’ to cover some interesting properties of complex numbers.

If there is continued interest in topics like these, I would be happy to cover them in future essays. So, please be sure to let me know in the comments below.

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Credit: My work on this essay was inspired by the work done by Jens Fehlau.

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Further reading that might interest you: What Is So Special About 69!? and How To Actually Subtract Using Addition?