The antiderivative of 1/(x^4+1): Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my series on the computation of

$\displaystyle \int \frac{dx}{x^4+1}$

Part 1: Introduction.

Part 2: Factoring the denominator using De Moivre’s Theorem.

Part 3: Factoring the denominator using the difference of two squares.

Part 4: The partial fractions decomposition of the integrand.

Part 5: Partial evaluation of the resulting integrals.

Part 6: Evaluation of the remaining integrals.

Part 7: An apparent simplification using a trigonometric identity.

Part 8: Discussion of the angles for which the identity holds.

Part 9: Proof of the angles for which the identity holds.

Part 10: Implications for using this identity when computing definite integrals.

The antiderivative of 1/(x^4+1): Part 2

This antiderivative has arguable the highest ratio of “really hard to compute” to “really easy to write”:

$\displaystyle \int \frac{1}{x^4 + 1} dx$

To compute this integral, I will use the technique of partial fractions. This requires factoring the denominator over the real numbers, which can be accomplished by finding the roots of the denominator. In other words, I need to solve

$x^4 + 1 = 0$ ,

$z^4 = -1$ .

I switched to the letter $z$ since the roots will be complex. The four roots of this quartic equation can be found with De Moivre’s Theorem by writing

$z = r (\cos \theta + i \sin \theta)$ ,

where $r$ is a real number, and

$-1 + 0i = 1(\cos \pi + \i \sin \pi)$

By De Moivre’s Theorem, I obtain

$r^4 (\cos 4\theta + i \sin 4 \theta) = 1 (\cos \pi + i \sin \pi)$ .

Matching terms, I obtain the two equations

$r^4 = 1$ and $4\theta = \pi + 2\pi n$

$r = 1$ and $\theta = \displaystyle \frac{\pi}{4} + \displaystyle \frac{\pi n}{2}$

$r = 1$ and $\theta = \displaystyle \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$ .

This yields the four solutions

$z = 1 \left[ \cos \displaystyle \frac{\pi}{4} + i \sin \frac{\pi}{4} \right] = \displaystyle \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}$

$z = 1 \left[ \cos \displaystyle \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right] = -\displaystyle \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}$

$z = 1 \left[ \cos \displaystyle \frac{5\pi}{4} + i \sin \frac{5\pi}{4} \right] = -\displaystyle \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}$

$z = 1 \left[ \cos \displaystyle \frac{7\pi}{4} + i \sin \frac{7\pi}{4} \right] = \displaystyle \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}$

Therefore, the denominator $x^4 + 1$ can be written as the following product of linear factors over the complex plane:

$\displaystyle \left(x - \left[ \displaystyle \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right] \right)\left(x - \left[ \displaystyle \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \right] \right) \left(x - \left[ -\displaystyle \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right] \right) \left(x - \left[ - \displaystyle \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \right] \right)$

$\displaystyle \left(\left[x - \displaystyle \frac{\sqrt{2}}{2} \right] - i \frac{\sqrt{2}}{2} \right)\left( \left[ x - \displaystyle \frac{\sqrt{2}}{2} \right] + i \frac{\sqrt{2}}{2} \right) \left( \left[ x + \displaystyle \frac{\sqrt{2}}{2} \right] - i \frac{\sqrt{2}}{2} \right) \left( \left[ x + \displaystyle \frac{\sqrt{2}}{2} \right] + i \frac{\sqrt{2}}{2} \right)$

$\displaystyle \left(\left[x - \displaystyle \frac{\sqrt{2}}{2} \right]^2 - \left[ i \frac{\sqrt{2}}{2} \right]^2 \right) \left( \left[ x + \displaystyle \frac{\sqrt{2}}{2} \right]^2 - \left[i \frac{\sqrt{2}}{2} \right]^2 \right)$

$\displaystyle \left(x^2 - x \sqrt{2} + \displaystyle \frac{1}{2} + \displaystyle \frac{1}{2}\right) \left(x^2 + x \sqrt{2} + \displaystyle \frac{1}{2} + \displaystyle \frac{1}{2}\right)$

$\displaystyle \left(x^2 - x \sqrt{2} + 1 \right) \left(x^2 + x \sqrt{2} + 1\right)$ .

We have thus factored the denominator over the real numbers:

$\displaystyle \int \frac{dx}{x^4 + 1} = \displaystyle \int \frac{dx}{\left(x^2 - x \sqrt{2} + 1 \right) \left(x^2 + x \sqrt{2} + 1\right)}$ ,

and the technique of partial fractions can be applied.

I’ll continue the calculation of this integral with tomorrow’s post.

How I Impressed My Wife: Part 5h

Earlier in this series, I gave three different methods of showing that

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.$

Using the fact that $Q$ is independent of $a$ , I’ll now give a fourth method.

Since $Q$ is independent of $a$ , I can substitute any convenient value of $a$ that I want without changing the value of $Q$ . As shown in previous posts, substituting $a =0$ yields the following simplification:

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + 2 \cdot 0 \cdot \sin x \cos x + (0^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\pi}^{\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{u^4 + (4 b^2 - 2) u^2 + 1}$

The four roots of the denominator satisfy

$u^2 = \displaystyle 1 - 2b^2 \pm 2|b| \sqrt{b^2 - 1}$

So far, I’ve handled the cases $|b| = 1$ and $|b| > 1$ . In today’s post, I’ll start considering the case $|b| < 1$ .

Factoring the denominator is a bit more complicated if $|b| < 1$ . Using the quadratic equation, we obtain

$u^2 = \displaystyle 1 - 2b^2 \pm 2|b| i \sqrt{1-b^2}$

However, unlike the cases $|b| \ge 1$ , the right-hand side is now a complex number. So, To solve for $u$ , I’ll use DeMoivre’s Theorem and some surprisingly convenient trig identities. Notice that

$(1-2b^2)^2 + (2|b| \sqrt{1-b^2})^2 = 1 - 4b^2 + b^4 + 4b^2 (1 - b^2) = 1 - 4b^2 + b^4 + 4b^2 - b^4 = 1$ .

Therefore, the four complex roots of the denominator satisfy $|u^2| = 1$ , or $|u| = 1$ . This means that all four roots can be written in trigonometric form so that

$u^2 = \cos 2\phi + i \sin 2\phi$ ,

where $2\phi$ is some angle. (I chose the angle to be $2\phi$ instead of $\phi$ for reasons that will become clear shortly.)

I’ll begin with solving

$u^2 = \displaystyle 1 - 2b^2 + 2|b| i \sqrt{1-b^2}$ .

Matching the real and imaginary parts, we see that

$\cos 2\phi = 1-2b^2$ ,

$\sin 2\phi = 2|b| \sqrt{1-b^2}$

This completely matches the form of the double-angle trig identities

$\cos 2\phi = 1 - 2\sin^2 \phi$ ,

$\sin 2 \phi = 2 \sin\phi \cos \phi$ ,

and so the problem reduces to solving

$u^2 = \cos 2\phi + i \sin 2\phi$ ,

where $\sin \phi = |b|$ and $\cos \phi = \sqrt{1-b^2}$. By De Moivre’s Theorem, I can conclude that the two solutions of this equation are

$u = \pm(\cos \phi + i \sin \phi)$ ,

$u = \pm( \sqrt{1-b^2} + i |b|)$ .

I could re-run this argument to solve $u^2 = \displaystyle 1 - 2b^2 - 2|b| i \sqrt{1-b^2}$ and get the other two complex roots. However, by the Conjugate Root Theorem, I know that the four complex roots of the denominator $u^4 + (4 b^2 - 2) u^2 + 1$ must come in conjugate pairs. Therefore, the four complex roots are

$u = \pm \sqrt{1-b^2} \pm i |b|$ .

Therefore, I can factor the denominator as follows:

$u^4 + (4 b^2 - 2) u^2 + 1 = (u - [\sqrt{1-b^2} + i|b|])(u - [\sqrt{1-b^2} - i|b|])$

$\qquad \times (u - [-\sqrt{1-b^2} + i|b|])(u - [-\sqrt{1-b^2} + i|b|])$

$= (u - \sqrt{1-b^2} - i|b|)(u - \sqrt{1-b^2} + i|b])$

$\qquad \times (u +\sqrt{1-b^2} + i|b|)(u +\sqrt{1-b^2} + i|b|)$

$= ([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)$

To double-check my work, I can directly multiply this product:

$([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)$

$= (u^2 - 2u \sqrt{1-b^2} + 1 - b^2 + b^2) (u^2 + 2u \sqrt{1-b^2} + 1 - b^2 + b^2)$

$= ([u^2 +1] - 2u\sqrt{1-b^2})([u^2+1] + 2u\sqrt{1-b^2})$

$= [u^2+1]^2 - [2u\sqrt{1-b^2}]^2$

$= u^4 + 2u^2 + 1 - 4u^2 (1-b^2)$

$= u^4 + u^2 (2 - 4[1-b^2]) + 1$

$= u^4 + u^2 (4b^2 - 2) + 1$ .

So, at last, I can rewrite the integral $Q$ as

$Q = \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{ ([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)}$

I’ll continue with this fourth evaluation of the integral, continuing the case $|b| < 1$ , in tomorrow’s post.

Calculators and Complex Numbers: Index

I’m using the Twelve Days of Christmas (and perhaps a few extra days besides) to do something that I should have done a long time ago: collect past series of posts into a single, easy-to-reference post. The following posts formed my series on how the trigonometric form of complex numbers, DeMoivre’s Theorem, and extending the definitions of exponentiation and logarithm to complex numbers.

Part 1: Introduction: using a calculator to find surprising answers for $\ln(-5)$ and $\sqrt[3]{-8}$ . See the video below.

Part 2: The trigonometric form of complex numbers.

Part 3: Proving the theorem

$\left[ r_1 (\cos \theta_1 + i \sin \theta_1) \right] \cdot \left[ r_2 (\cos \theta_2 + i \sin \theta_2) \right] = r_1 r_2 (\cos [\theta_1+\theta_2] + i \sin [\theta_1+\theta_2])$

Part 4: Proving the theorem

$\displaystyle \frac{ r_1 (\cos \theta_1 + i \sin \theta_1) }{ r_2 (\cos \theta_2 + i \sin \theta_2) } = \displaystyle \frac{r_1}{r_2} (\cos [\theta_1-\theta_2] + i \sin [\theta_1-\theta_2])$

Part 5: Application: numerical example of De Moivre’s Theorem.

Part 6: Proof of De Moivre’s Theorem for nonnegative exponents.

Part 7: Proof of De Moivre’s Theorem for negative exponents.

Part 8: Finding the three cube roots of -27 without De Moivre’s Theorem.

Part 9: Finding the three cube roots of -27 with De Moivre’s Theorem.

Part 10: Pedagogical thoughts on De Moivre’s Theorem.

Part 11: Defining $z^q$ for rational numbers $q$ .

Part 12: The Laws of Exponents for complex bases but rational exponents.

Part 13: Defining $e^z$ for complex numbers $z$

Part 14: Informal justification of the formula $e^z e^w = e^{z+w}$ .

Part 15: Simplification of $e^{i \theta}$ .

Part 16: Remembering DeMoivre’s Theorem using the notation $e^{i \theta}$ .

Part 17: Formal proof of the formula $e^z e^w = e^{z+w}$ .

Part 18: Practical computation of $e^z$ for complex $z$ .

Part 19: Solving equations of the form $e^z = w$ , where $z$ and $w$ may be complex.

Part 20: Defining $\log z$ for complex $z$ .

Part 21: The Laws of Logarithms for complex numbers.

Part 22: Defining $z^w$ for complex $z$ and $w$ .

Part 23: The Laws of Exponents for complex bases and exponents.

Part 24: The Laws of Exponents for complex bases and exponents.

Calculators and complex numbers (Part 24)

In this series of posts, I explore properties of complex numbers that explain some surprising answers to exponential and logarithmic problems using a calculator (see video at the bottom of this post). These posts form the basis for a sequence of lectures given to my future secondary teachers.

To begin, we recall that the trigonometric form of a complex number $z = a+bi$ is

$z = r(\cos \theta + i \sin \theta) = r e^{i \theta}$

where $r = |z| = \sqrt{a^2 + b^2}$ and $\tan \theta = b/a$ , with $\theta$ in the appropriate quadrant. As noted before, this is analogous to converting from rectangular coordinates to polar coordinates.

Theorem. If $z = x + i y$ , where $x$ and $y$ are real numbers, then

$e^z = e^x (\cos y + i \sin y)$

Definition. Let $z = r e^{i \theta}$ be a complex number so that $-\pi < \theta \le \theta$ . Then we define

$\log z = \ln r + i \theta$ .

Definition. Let $z$ and $w$ be complex numbers so that $z \ne 0$ . Then we define

$z^w = e^{w \log z}$

Technical point: for the latter two definitions, these are the principal values of the functions. In complex analysis, these are usually considered multiply-defined functions. But I’m not going to worry about this technicality here and will only consider the principal values.

This is the last post in this series, where I state some generalizations of the Laws of Exponents for complex numbers.

In yesterday’s post, we saw that $z^{w_1} z^{w_2} = z^{w_1 + w_2}$ as long as $z \ne 0$ . This prevents something like $0^4 \cdot 0^{-3} = 0^1$ , since $0^{-3}$ is undefined.

Theorem. Let $z \in \mathbb{C} \setminus \{ 0 \}$ , $w \in \mathbb{C}$ , and $n \in \mathbb{Z}$ . Then $(z^w)^n = z^{wn}$ .

As we saw in a previous post, the conclusion could be incorrect outside of the above hypothesis, as $\displaystyle \left[ (-1)^3 \right]^{1/2} \ne (-1)^{3/2}$ .

Theorem. Let $u \in \mathbb{R}$ and $z \in \mathbb{C}$ . Then $(e^u)^z = e^{uz}$ .

Theorem. Let $x, y > 0$ be real numbers and $z \in \mathbb{C}$ . Then $x^z y^z = (xy)^z$ .

Again, the conclusion of the above theorem could be incorrect outside of these hypothesis, as $(-2)^{1/2} (-3)^{1/2} \ne \left[ (-2) \cdot (-3) \right]^{1/2}$ .

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For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures.

Calculators and complex numbers (Part 23)

To begin, we recall that the trigonometric form of a complex number $z = a+bi$ is

$z = r(\cos \theta + i \sin \theta) = r e^{i \theta}$

Theorem. If $z = x + i y$ , where $x$ and $y$ are real numbers, then

$e^z = e^x (\cos y + i \sin y)$

Definition. Let $z = r e^{i \theta}$ be a complex number so that $-\pi < \theta \le \theta$ . Then we define

$\log z = \ln r + i \theta$ .

Definition. Let $z$ and $w$ be complex numbers so that $z \ne 0$ . Then we define

$z^w = e^{w \log z}$

In the remaining posts in this series, I want to explore which properties of exponential functions remain true when complex numbers are used.

To begin, if $w$ is a real rational number, then there is an alternative definition of $z^w$ that matches De Moivre’s Theorem. Happily, the two definitions agree. Suppose that $z = r e^{i \theta}$ with $-\pi < \theta \le \pi$ . Then

$z^w = e^{w \log z}$

$= e^{w [\ln r + i \theta]}$

$= e^{w \ln r + i w \theta}$

$= e^{w \ln r} e^{i w \theta}$

$= r^w (\cos w\theta + i \sin \theta)$

Next, one of the Laws of Exponents remains true even for complex numbers:

$z^{w_1} z^{w_2} = e^{w_1 \log z} e^{w_2 \log z}$

$= e^{w_1 \log z + w_2 \log z}$

$= e^{(w_1 + w_2) \log z}$

$= z^{w_1 + w_2}$ .

However, in previous posts, we’ve seen that the rules $(x^y)^z = x^(yz)$ and $x^z y^z = (xy)^z$ may not be true if nonpositive bases, let alone complex bases, are used.

We can also derive the usual rules $z^0 = 1$ and $z^{-w} = \displaystyle \frac{1}{z^w}$ . First,

$z^0 = e^{0 \log z} = e^0 = 1$ .

Next, we think like an MIT freshman and use the above Law of Exponents to observe that

$z^w z^{-w} = z^{w-w} = z^0 = 1$ .

Dividing, we see that $z^{-w} = \displaystyle \frac{1}{z^w}$ .

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For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures.

Calculators and complex numbers (Part 16)

To begin, we recall that the trigonometric form of a complex number $z = a+bi$ is

$z = r(\cos \theta + i \sin \theta) = r e^{i \theta}$

Real mathematicians use the notation $e^{i \theta}$ to represent $\cos \theta + i \sin \theta$ . I say this because I’ve seen textbooks that basically invented the non-standard notation $\hbox{cis} \, \theta$ (pronounced siss), where presumably the c represents $\cos$ and the s represents $\sin$ . I express my contempt for this non-standard notation by saying that this is a sissy way of writing it.

With this shorthand notation of $r e^{i \theta}$ , several of the theorems that we’ve discussed earlier in this series of posts become a lot more memorable.

First, the formula

$\left[ r_1 ( \cos \theta_1 + i \sin \theta_1) \right] \cdot \left[ r_2 ( \cos \theta_2 + i \sin \theta_2 ) \right] = r_1 r_2 \left[ \cos( \theta_1 + \theta_2) + i \sin (\theta_1 + \theta_2) \right]$

can be rewritten as something that resembles the familiar Law of Exponents:

$r_1 e^{i \theta_1} r_2 e^{i \theta_2} = r_1 r_2 e^{i(\theta_1 + \theta_2)}$

Similarly, the formula

$\displaystyle \frac{ r_1 ( \cos \theta_1 + i \sin \theta_1)}{ r_2 ( \cos \theta_2 + i \sin \theta_2 ) } = \displaystyle \frac{r_1}{ r_2} \left[ \cos( \theta_1 - \theta_2) + i \sin (\theta_1 - \theta_2) \right]$

can be rewritten as

$\displaystyle \frac{r_1 e^{i \theta_1}}{ r_2 e^{i \theta_2}} = \displaystyle \frac{r_1 }{r_2} e^{i(\theta_1 - \theta_2)}$

Finally, DeMoivre’s Theorem, or

$\left[ r (\cos \theta + i \sin \theta) \right]^n = r^n (\cos n \theta + i \sin n \theta)$

can be rewritten more comfortably as

$\left( r e^{i \theta} \right)^n = r^n e^{i n \theta}$

When showing these to students, I stress that these are not the formal proofs of these statements… the formal proofs required trig identites and mathematical induction, as shown in previous posts. That said, now that the proofs have been completed, the $e^{i \theta}$ notation provides a way of remembering these formulas that wasn’t immediately obvious when we began this unit on the trigonometric form of complex numbers.

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For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures.

Calculators and complex numbers (Part 12)

To begin, we recall that the trigonometric form of a complex number $z = a+bi$ is

$z = r(\cos \theta + i \sin \theta)$

There’s a shorthand notation for the right-hand side ( $r e^{i \theta}$ ) that I’ll justify later in this series.

In the previous post, we made the following definition for $z^q$ if $q$ is a rational number and $-\pi < \theta \le \pi$ . (Technically, this is the definition for the principal root.)

Definition. $z^q = r^q (\cos q \theta + i \sin q \theta)$ .

As it turns out, one of the usual Laws of Exponents remains true even if complex numbers are permitted.

Theorem. $z^{q_1} z^{q_2} = z^{q_1 + q_2}$

Proof. Using the rule for multiplying complex numbers that are in trigonometric form:

$z^{q_1} z^{q_2} = r^{q_1} (\cos q_1 \theta + i \sin q_1 \theta) \cdot r^{q_2} (\cos q_2\theta + i \sin q_2 \theta)$

$= r^{q_1+q_2} ( \cos [q_1 \theta +q_2\theta] + i \sin [q_1\theta +q_2 \theta])$

$= r^{q_1+q_2} ( \cos [q_1+q_2]\theta + i \sin [q_1+q_2] \theta)$

$= z^{q_1+q_2}$

However, other Laws of Exponents no longer are true. For example, it may not be true that $(zw)^q$ is equal to $z^q w^q$ . My experience is that this next example is typically presented in secondary schools at about the time that the number $i$ is first introduced. Let $z = -2$ , $w = -3$ , and $q = 1/2$ . Then

$\sqrt{-2} \cdot \sqrt{-3} = i \sqrt{2} i \sqrt{3} = -\sqrt{6} \ne \sqrt{6} = \sqrt{(-2) \cdot (-3)}$ .

Furthermore, the expression $(z^{q_1})^{q_2}$ does not have to equal $z^{q_1 q_2}$ if $z$ is complex. Let $z = -1$ , $q_1 = 3$ , and $q_2 = 1/2$ . Then

$\left[ (-1)^3 \right]^{1/2} = (-1)^{1/2}$

$= [1 (\cos \pi + i \sin \pi)]^{1/2}$

$= \displaystyle 1^{1/2} \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right)$

$= 1(0+1i)$

$= i$ .

However,

$(-1)^{3/2} = [1 (\cos \pi + i \sin \pi)]^{3/2}$

$= \displaystyle 1^{3/2} \left( \cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2} \right)$

$= 1(0-1i)$

$= -i$ .

All this to say, the usual Laws of Exponents that work for real exponents and positive bases don’t have to work if the base is permitted to be complex… or even negative.

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For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures.

Calculators and complex numbers (Part 11)

In today’s post, at long last, I can explain one of the unexpected results of the calculator shown in the opening sections of the video below: the different answers for $(-8)^{1/3}$ and $(-8+0i)^{1/3}$ .

To begin, we recall that the trigonometric form of a complex number $z = a+bi$ is

$z = r(\cos \theta + i \sin \theta)$

There’s a shorthand notation for the right-hand side ( $r e^{i \theta}$ ) that I’ll justify later in this series.

In previous posts, we discussed De Moivre’s Theorem:

Theorem. If $n$ is an integer, then $z^n = \left[ r (\cos \theta + i \sin \theta) \right]^n = r^n (\cos n \theta + i \sin n \theta)$ .

This motivates the following definition:

Definition. If $q$ is a rational number, then $z^q = r^q (\cos q \theta + i \sin q \theta)$ if $theta$ is chosen to be in the interval $-\pi < \theta \le \pi$ .

Technically speaking, this defines the principal value of $z^q$ ; however, for the purposes of this post, I’ll avoid discussion of branch cuts and other similar concepts from complex analysis. When presenting this to my future secondary teachers, I’ll often break the presentation by asking my students why it’s always possible to choose the angle $\theta$ to be in the range $(-\pi,\pi]$, and why it’s necessary to include exactly one of the two endpoints of this interval. I’ll also point out that this interval really could have been $[0,2\pi)$ or any other interval with length $2\pi$ , but we choose $(-\pi,\pi]$ for a very simple reason: tradition.

Using this definition, let’s compute $(-8+0i)^{1/3}$ . To begin,

$-8+0i = 8 (\cos \pi + i \sin \pi)$ .

So, by definition,

$(-8+0i)^{1/3} = 8^{1/3} \displaystyle \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right)$

$= \displaystyle 2 \left( \frac{1}{2} + \frac{\sqrt{3}}{2}i \right)$

$= 1 + i \sqrt{3}$

As noted in an earlier post in this series, this is one of the three solutions of the equation $z^3 = -8$ . Using De Moivre’s Theorem, the other two solutions are $z = -2$ and $z = 1 - i \sqrt{3}$ .

So, when $(-8+0i)^{1/3}$ is entered into the calculator, the answer $1 + i \sqrt{3}$ is returned.

On the other hand, when $(-8)^{1/3}$ is entered into the calculator, the calculator determines the solution that is a real number (if possible). So the calculator returns $-2$ and not $1 + i \sqrt{3}$ .

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For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures.

Calculators and complex numbers (Part 10)

To begin, we recall that the trigonometric form of a complex number $z = a+bi$ is

$z = r(\cos \theta + i \sin \theta)$

There’s a shorthand notation for the right-hand side ( $r e^{i \theta}$ ) that I’ll justify later in this series.

Today, I want to share some pedagogical thoughts about this series of posts. I’ll continue with the mathematical development of these ideas tomorrow.

My experience is that most math majors have never seen this particular application of trigonometry to find the $n$ th roots of complex numbers… or even are familiar with the idea of expressing a complex number into trigonometric form at all. This personally surprises me, as this was just one of the topics that I had to learn when I took Precalculus (which was called Trig/Analysis when I took it). I really don’t know if I was fortunate to be exposed to these ideas in my secondary curriculum of the 1980’s or if this was simply a standard topic back then. However, at least in Texas, the trigonometric form of complex numbers does not appear to be a standard topic these days.

This certainly isn’t the most important topic in the mathematics secondary curriculum. That said, I really wish that this was included in a standard Pre-AP course in Precalculus to better serve the high school students who are most likely to take more advanced courses in mathematics and science in college. These ideas are simply assumed in, say, Differential Equations, when students are asked to solve

$y^{5} – 32 y = 0$.

The characteristic equation of this differential equation is $r^5 - 32 = 0$ , and it’s really hard to find all five complex roots unless De Moivre’s Theorem is employed.

To give another example: In physics, even a cursory look at my old electricity and magnetism text reveals that familiarity with the trigonometric form of complex numbers can only facilitate student understanding of these physical concepts. Ditto for many concepts in electrical engineering. Stated another way, students who aren’t used to thinking of complex numbers in this way may struggle through physics and engineering in ways that could have been avoided with prior mathematical training.

green line

For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures.