On Negative Fractional Exponents

This question came in, and I wanted to feature it here as the first post in this category, because it touches on my favourite branch of math: Complex Analysis.

The essence of the question asked by David is this:

When you type (-7/3)^(-7/3) into most calculators, you end up with a complex number as a result. Shouldn’t the answer be real, as the cube root for negative numbers exists?

This question came in from a 9th grader with the presence of mind to actually think about what was being presented to them as an answer. Most people, unfortunately, do not do that and simply write down whatever their calculator tells them. The fact that they came up with this question shows the importance of thinking when solving mathematical problems. It is a very good trait to have, and I really hope they do keep questioning things because that is the key to learning.

So, let’s address the easy part of the question. Yes, the cube root for negative numbers is real and defined, and yes, (-7/3)^(-7/3) is real. But it still leaves the question: “Why are calculators providing me with a complex root?”. Checking the calculator’s answer by cubing the result will show that this value is, indeed, a root. So, what is going on?

We first need to introduce the Fundamental Theorem of Algebra. Sounds fancy and important, because it is! It says this:

Every non-zero, single-variable, degree n polynomial with complex coefficients has, counted with multiplicity, exactly n complex roots

Seems like a mouthful, but let’s break this down. It basically says that any polynomial of degree n has n complex roots. Note the usage of the word “complex” here. “Complex” does not mean it will have imaginary components. All numbers are complex numbers. Real numbers are also considered complex numbers. All this theorem says is that for a polynomial of degree n, you will always have n distinct roots. These roots may be real, may be purely imaginary, or may be a mix of both – hence, complex. Roots may also be duplicated (for example, a polynomial may have two roots, both at the same point, called its multiplicity, but that lies outside the scope of this post). A polynomial of degree 5 will have 5 roots, a polynomial of degree 50 will have 50 roots and so forth.

So, how does this apply to the problem at hand? When trying to take the root of a number (the type of root doesn’t really matter), we’re essentially solving for the roots of a polynomial. Consider the problem that started this whole discussion:

\(x= -(\frac{7}{3})^{-\frac{7}{3}}\\
x= -(\frac{3}{7})^{\frac{7}{3}}\\
x^3 = -(\frac{3}{7})^7\\
x^3 = -\frac{2187}{823543}\\
x^3 +\frac{2187}{823543} = 0\\
\text{…which is a polynomial of degree 3}

So now, we agree that we should have three “answers” to this. Three distinct roots. This is part #1 of the puzzle.

Now, we will introduce part 2 of the puzzle: what a root actually looks like, geometrically. It might be a little long, as I try to set up the foundations for what is going on, but bear with me!

We first need to talk about the complex plane. Just as a regular number line which is 1-dimensional (which goes left and right) can be used to represent real numbers, we can use a 2-dimensional plane to represent complex numbers, very much like a co-ordinate system. The x-coordinate represents the real part of the number and the y-coordinate represents the imaginary part. For example, if we wanted to represent 3 + 4i on the complex plane, we’d go 3 units to the right of the origin, then go up 4 units and put a dot over there.

When we express complex numbers in terms of their real component and their imaginary component, we call it rectangular form because it looks like we’re specifying the sides of a rectangle. But, there is another way to express these numbers. We can specify the length of the line, r, and then say what angle it makes with the x axis, θ.

In this case, r would be 5 (calculated via Pythagorean theorem, r being the hypotenuse) and θ would be about 53.13° (calculated via the inverse tangent). We can write this as 5 cis(53.13°) or 5(cos(53.13) + isin(53.13)).  But the cis() function is well-known, and Leonard Euler had another way of writing it, called “Euler’s formula”:

\(cis (x) = \cos (x) + i \sin (x) = e^{ix}\)

So, to sum up, we have a number of ways of writing complex numbers. These are:

\(\begin{align*}3 + 4i  &= \; \text{(Rectangular form)}\\
5  \text{cis}(53.13) &= \; \text{(Cis form)}\\
5(\cos (53.13) + i \sin (53.13) &= \; \text{(Polar form)}\\
5e^{0.93i} &= \; \text{(Exponential form)}\end{align*}

(the number inside the exponential form is different because you have to convert to radians for this equation to work)

Now, let’s take a look at the exponential form of a complex number. Particularly, what happens when you want to take an n-th root.

\((re^{i \theta})^n = r_0 e^{i \theta _0}\\
r^ne^{in \theta} = r_0 e^{i \theta _0}

First, take a look at the relation between r and r0. The value of r0 is simply the nth root of r. This number is always defined because r is always positive.

Now, take a look at the relation between the exponential functions. Essentially, one is a scaling (of factor n of the other). So, when we take roots, we are basically multiplying that angle by a certain factor.

And geometrically, that is what we are doing when taking roots. We draw a circle (radius r) centered at the origin for any complex number. When we want to take a root, we split that circle into a number of sections of the root’s degree. Those sections define what the roots are.

For example, let’s look at  our 3+4i example. The cube roots look like this (the red dots are the points that denote the value of the root):

The 4th roots look like this:

..and to really go wild, the 8th roots look like this:

And now, finally, we are ready to address the question at hand. What are the roots of (-7/3)^(-7/3)? The roots look like this:

You can see that it has two imaginary roots, and one real root (the root that you would expect to get). In fact, the roots of this are:

\(\frac{9}{49} \sqrt[3]{\frac{3}{7}} e^{-i \frac{\pi}{3}} \approx 0.0692 – 0.1199i\\
-\frac{9}{49} \sqrt[3]{\frac{3}{7}} \approx -0.1385\\
\frac{9}{49} \sqrt[3]{\frac{3}{7}} e^{i \frac{\pi}{3}} \approx 0.0692 + 0.1199i\\\)

Now, the last part of the puzzle is a question as to why the calculator returns the imaginary root, rather than the real root. Most algorithms (when programmed correctly) will provide you with what is called the Principal Root. In essence, the principal root is the root that has the smallest positive angle between it and the positive x-axis (going counter-clockwise). In the diagram above, you can see that there is a root between the real root that you would expect to be displayed, and that is the root that will be shown to you.

And hopefully that should fully answer the question. Got any other questions relating to this? Let me know in the comments and I’ll try to expand further.

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2 Responses to On Negative Fractional Exponents

  1. romeet.M says:

    i love this answer , very thoroughly explained and i was never introduced to the way a root of a complex number can be shown geometrically , even though i learned about complex roots , and you have gone deep enough without having said that “you will learn about this in higher classes”.And you did all of this just for a comment on khan’s differential equation vid , i appreciate this

    • aommaster says:


      Thank you very much. Yes, unfortunately, this ends up being taught in university for two reasons:
      1. It is part of Complex Analysis, an advanced level of calculus and is only used in very specialized fields
      2. It would take too much time to fit into a school syllabus

      But as a result of that, I believe school students miss out on the elegance behind the math, a common theme you will notice if you take math to a higher level. Things that were just taken for granted like identities actually make a lot of sense.

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