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In high school we learn that the cube root of $-8$ is $-2$. Much later some of us learn about the single valued natural logarithm of a complex number, and that $w^z = e^{z\cdot Lz(w)}$ when $w$ and $z$ are complex. That gives a different result for the cube root of $-8$. When would what we learn in high school not serve an undergraduate majoring in physics or engineering?

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  • $\begingroup$ When do they have problems? The first time they do the calculation on their calculator. $\endgroup$ – Gerald Edgar Sep 14 '20 at 1:45
  • $\begingroup$ I don't see this as much of a concern, at least in the U.S. First, high school students will typically have seen and used De Moivre's theorem in trigonometry (and also Euler's forumula in a 2nd semester first year calculus class (when Taylor series is covered), and better students will know how to factor $x^3 + 8$ as a sum of cubes along with knowing how to use the quadratic formula when non-real roots are involved. (continued) $\endgroup$ – Dave L Renfro Sep 14 '20 at 12:21
  • $\begingroup$ Second, physics students (and the more math-intensive engineering students, such as mechanical and electrical enginnering) will take an upper level mathematical methods course that includes some complex variables (using something like Kreyzig's book or Kaplan's book). $\endgroup$ – Dave L Renfro Sep 14 '20 at 12:21
  • $\begingroup$ Am I misreading or misunderstanding? What "single-valued natural log of a complex number" is there? Someone's convention? Or is that not what was intended? $\endgroup$ – paul garrett Sep 14 '20 at 23:23
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    $\begingroup$ @paulgarrett: Consistent with the comment by Daniel Collins, the single-valued natural log of a complex number is in eq 4.2.2 and 4.2.3 at dlmf.nist.gov/4.2 $\endgroup$ – Ted Ersek Sep 15 '20 at 16:26
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This is a complicated question, and there are a number of articles written on the topic in the math education literature. Here are some of the entries that I would recommend (taken from the bottom of this answer):

  • Goel, Sudhir K., and Michael S. Robillard. "The Equation: $-2 = (-8)^\frac{1}{3} = (-8)^\frac{2}{6} = [(-8)^2]^\frac{1}{6} = 2$." Educational Studies in Mathematics 33.3 (1997): 319-320.

  • Tirosh, Dina, and Ruhama Even. "To define or not to define: The case of $(-8)^\frac{1}{3}$." Educational Studies in Mathematics 33.3 (1997): 321-330.

  • Choi, Younggi, and Jonghoon Do. "Equality Involved in 0.999... and $(-8)^\frac{1}{3}$" For the Learning of Mathematics 25.3 (2005): 13-36.

  • Woo, Jeongho, and Jaehoon Yim. "Revisiting 0.999... and $(-8)^\frac{1}{3}$ in School Mathematics from the Perspective of the Algebraic Permanence Principle." For the Learning of Mathematics 28.2 (2008): 11-16.

As the OP alludes to, there are in fact different definitions of the (single-valued) principal root in the real and complex contexts. It's quite common on SE for there to be confusion, cross-talk, and frustration between specialists in the two different domains, each assuming a different context and definition. I suppose that interacting with any of those types of discussions could count as an example for the OP's purpose.

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    $\begingroup$ Very good answer and comments above. I am an electrical engineer and find it most disappointing that we used complex numbers all the time in college, but neither my professors or text books explained when the rules for exponents that we learn in high school can be used when complex numbers are involved! They never told students a^c b^c = (a b)^c whenever both sides are defined, They don't tell us that a^c* b^(-c) = a^c/(b^c) for all a,b,c. Continued $\endgroup$ – Ted Ersek Sep 15 '20 at 17:12
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    $\begingroup$ And they don't tell us when (a^c)*(b^c) = (a b)^c, when (a^c)/(b^c) = (a/b)^c, (a^b)^c = a^(bc), when a^c*(1/b)^c = (a/b)^c, or when a^cb^(-c)=(a^c)/(b^c)=a^c*(1/b)^c=(a/b)^c. Since then, I have used Mathematica to figure out all of that and it's interesting, but I can't find it anywhere in print! $\endgroup$ – Ted Ersek Sep 15 '20 at 17:13
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That gives a different result for the cube root of -8.

It doesn't give a different result, it just gives two additional roots that are complex numbers, for a total of three roots.

A physics or engineering student in the US probably first learns about complex numbers in high school, but never sees any interesting applications. Then in college classes they get such applications (possibly not until they get to upper-division classes), and then they do have to do a certain amount of unlearning of facts that hold for the reals but not the complex numbers. (This unlearning should of course have happened in high school when they first saw complex numbers, but it doesn't really.) These applications would include, for example, analyzing linear filters in an electrical engineering class.

In my experience the particular issue you ask about has never been a problem for my physics students. The main problem with their high school background is that they have been drilled endlessly in the cartesian representation, so they have trouble transitioning to the polar representation, which is more natural and convenient for the applications we do. For example, it takes them 10 minutes to compute $1/i$, because they're doing it using some algorithm they've been taught for computing $(a+bi)/(c+di)$, rather than just visualizing $1/i$ on the unit circle.

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    $\begingroup$ I'll differ with, "It doesn't give a different result". The principal root actually does have different definitions in the real and complex contexts. E.g., the complex formula the OP gives is single-valued ("single valued natural logarithm"). This is recognized as a common problem in the literature, e.g., references at the end of this question: math.stackexchange.com/questions/1628759/… $\endgroup$ – Daniel R. Collins Sep 13 '20 at 13:58
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    $\begingroup$ @DanielR.Collins: AFAICT the choice of a principal root is not standardized, nor is it AFAIK a concept that has much use in physics and engineering applications. In these applications, we generally care about all the roots. E.g., in a diffeq like 7y''+y'-2=0, we want both solutions. Stuff like branch cuts is important if you're going to do complex analysis and contour integrals, but if physics and engineering students encounter that at all, it would typically be in grad school. The question is about undergrad physics and engineering. $\endgroup$ – Ben Crowell Sep 13 '20 at 18:48
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    $\begingroup$ @Daniel R. Collins: FYI, in my last 2 years' low-paid part time contract work, I've reviewed a few thousand items for a certain non-U.S. standardized test (and have provided additional support in other ways), and in some of the items involving logarithms (but not all items involving logarithms) "Log" (capital 'L') is used. They didn't realize this could be problematic for some test takers. $\endgroup$ – Dave L Renfro Sep 14 '20 at 10:44
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    $\begingroup$ Apart from the fantasies of test-manufacturing corporations, I really do not think there is any reliable, robust, durable notion of "principal log" or "principal cube root". Even if we want to pretend that there is a "standard choice", there are at least two... oops... The whole idea of dictating to mathematics that it adhere to human conventions is a bit silly, and really obnoxious if we examine students on conventions if that displaces asking about reality. Still, yes, I know, there are standardized tests... $\endgroup$ – paul garrett Sep 14 '20 at 23:03
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    $\begingroup$ Daniel R. Collins had the interpretation I intended. I am an electrical engineer and as an undergraduate I learned De Moiver's theorem and Euler's identity. Sadly, my professors and text books didn't explicitly state when the rules of exponents learned in high school can be applied when some numbers are complex numbers, but it seems undergraduates don't need to know how to compute w^z for complex values. $\endgroup$ – Ted Ersek Sep 15 '20 at 16:45
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The equation $(-8)^{1/3}=-2$ in isolation is taught in early algebra. Later, in precalculus, on learns about the Fundamental Theorem of Algebra. At this point, one starts to understand that this equation should be seen more generally in the context of roots of the polynomial $x^3+8=0$. One learns there are three roots, and falls back on the fact that $(-2)^3=-8$ to find one such root. Using the factor theorem and polynomial division with $\frac{x^3+8}{x+2}$, one reduces the problem of finding the other two roots to using the quadratic equation to solve $x^2-2x+4=0$. Typically these two solutions in precalculus are left in the form $x=1\pm i\sqrt{3}$. This is where many students stop developmentally, as described in Ben Crowell's answer.

Students going further in fields such as physics, engineering, and mathematics will develop further mathematically, and become comfortable with the polar representation of complex numbers. At this stage, the student will see the solutions of $z^3+8=0$ best represented by $z=2\exp\left(\frac{(2k+1)i\pi}3\right)$ with $k=0,1,2$.

So as others have pointed out in comments and answers, teaching that $(-8)^{1/3}=-2$ in elementary algebra is not going to be a problem for those who take more mathematics to become physicists, engineers, and mathematicians. In fact, teaching this arithmetic fact is the basic foundation for a future higher level understanding.

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When would what we learn in high school not serve an undergraduate majoring in physics or engineering?

It would begin to be a problem when you enter any field where complex numbers appear. Obvious places where complex numbers appear would be in linear ODEs, Fourier transforms, etc. In quantum mechanics, wave functions are complex valued.

In many of these situations, making the cube root single valued would mean that you miss real valued solutions so some equations. Here's an analogue to your example: $$y'''=-8y$$ As it turns out, the solutions are generated by those of the form $y=e^{rx}$ where $r$ is some constant. (I believe physicists call this an ansatz.) When you plug this form into the ODE, you get the requirement that $r^3=-8$. If you take the real valued solution you get $y=e^{-2x}$.

That misses the other two real valued generators: $$ y= e^{x} \cos\left(\sqrt{3}x\right) \quad \text{and} \quad y= e^{x} \sin\left(\sqrt{3}x\right)$$ which are formed from the complex root solutions: $y=e^{2 \omega_3 x}$, $y=e^{2 \omega_3^2 x}$ where $\omega_3 = e^{i\pi/3}=\cos(\pi/3)+i\sin(\pi/3)=\frac{1+i\sqrt{3}}{2}$.

If you only had the real root, you would think that the system rapidly settles down to $0$. However, while this is possible, the generic behavior of a solution is an increasingly wild oscillation.

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    $\begingroup$ I think students should be taught that we almost always deal with single valued function because the are much more useful than multi-valued functions in applications. Rather than talk about multi-valued functions the differential equation above should be solved by finding all solutions to z^3 = -8, and then you show them how to find those solutions. You could also have them verify that they really are solutions to z^3 = -8. $\endgroup$ – Ted Ersek Sep 15 '20 at 16:19

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