Calculators and complex numbers (Part 20)

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.

Over the past few posts, we developed the following theorem for computing e^z in the case that z is a complex number.

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

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

As a consequence, there are infinitely many complex solutions of the equation

e^z = -2 - 2i,

namely, z = \ln 2\sqrt{2} - \displaystyle \frac{3\pi}{4} + 2 \pi n i.

Choosing the solution that has an imaginary part in the interval (-\pi,\pi] leads to the definition of the complex logarithm.

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.

Of course, this looks like what the definition ought to be if one formally applies the Laws of Logarithms to r e^{i \theta}. So, for example,

\log (-2-2i) = \ln 2\sqrt{2} - \displaystyle \frac{3\pi}{4}

A technicality: this is the principal value of the complex logarithm. In complex analysis, this is technically thought of as a multiply-defined function.

The complex version of the natural logarithm function matches the ordinary definition when applied to real numbers. For example,

\log 6 = \log \left( 6 e^{0i} \right) = \ln 6 + 0 i = \ln 6.

A couple of observations. In high school, the symbol \log is usually dedicated to base 10. However, in higher-level mathematics courses, \log always means natural logarithm. That’s because, for the purposes of abstract mathematics, base-10 logarithms are practically useless. They are helpful for us people since our number system uses base 10; it’s easy for me to estimate \log_{10} 9000, but \ln 9000 requires a little more thought. But nearly all major theorems that involve logarithms specifically employ natural logarithms. Indeed, when I first become a professor, I had to remind myself that my students used \ln for natural logarithms and not \log. Still, I write \log_{10} for base-10 logarithms and not \log as a silent acknowledgment of the use of the symbol in higher-level courses.

This use of the logarithm explains the final results of the calculator in the video below. When \ln(-5) is entered, it assumes that a real answer is expected, and so the calculatore returns an error message. On the other hand, when \ln(-5+0i) is entered, it assumes that the user wants the principal complex logarithm. Since -5+0i = 5 e^{i \pi}, the calculator correctly returns \ln 5 + \pi i as the answer. (Of course, the calculator still uses \ln and not \log to mean natural logarithm.)

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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 19)

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.

Over the past few posts, we developed the following theorem for computing e^z in the case that z is a complex number.

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

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

Example. Find all complex numbers z so that e^z = 5.

Solution. If z = x + iy, then

e^x (\cos y + i \sin y) = 5 (\cos 0 + i \sin 0)

Matching parts, we see that e^x = 5 and that the angle y must be coterminal with 0 radians. In other words,

x = \ln 5 \qquad and \qquad y = 2\pi n for any integer n.

Therefore, there are infinitely many answers: z = \ln 5 + 2 \pi n i.

Notice that there’s nothing particularly special about the number 5. This could have been any nonzero number, including complex numbers, and there still would have been an infinite number of solutions. (This is completely analogous to solving a trigonometric equation like \sin \theta = 1, which similarly has an infinite number of solutions.) For example, the complex solutions of the equation

e^z = -2 - 2i

are z = \ln 2\sqrt{2} - \displaystyle \frac{3\pi}{4} + 2 \pi n i.

These observations lead to the following theorems, which I’ll state without proof.

Theorem. The range of the function f(z) = e^z is \mathbb{C} \setminus \{ 0 \}.

Theorem. e^z = e^w \Longleftrightarrow z = w + 2\pi n i.

Naturally, these conclusions are different than the normal case when z is assumed to be a real number.

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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 18)

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.

Definition. If z is a complex number, then we define

e^z = \displaystyle \sum_{n=0}^{\infty} \frac{z^n}{n!}

This of course matches the Taylor expansion of e^x for real numbers x.

In the last few posts, we proved the following theorem.

Theorem. If z and w are complex numbers, then $e^z e^w = e^{z+w}$.

This theorem allows us to compute e^z without directly plugging into the above infinite series.

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

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

Proof. With the machinery that’s been developed over the past few posts, this one is actually a one-liner:

e^z = e^{x+iy} = e^x e^{iy} = e^x (\cos y + i \sin y).

For example,

e^{4+\pi i} = e^4 (\cos \pi + i \sin \pi) = -e^4

Notice that, with complex numbers, it’s perfectly possible to take e to a power and get a negative number. Obviously, this is impossible when using only real numbers.

Another example:

e^{-2+3i} = e^{-2} (\cos 3 +i \sin 3)

In this answer, we have to remember that the angle is 3 radians and not 3 degrees.

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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 17)

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.

Definition. If z is a complex number, then we define

e^z = \displaystyle \sum_{n=0}^{\infty} \frac{z^n}{n!}

Even though this isn’t the usual way of defining the exponential function for real numbers, the good news is that one Law of Exponents remains true. (At we saw in an earlier post in this series, we can’t always assume that the usual Laws of Exponents will remain true when we permit the use of complex numbers.)

Theorem. If z and w are complex numbers, then e^z e^w = e^{z+w}.

In yesterday’s post, I gave the idea behind the proof… group terms where the sums of the exponents of z and w are the same. Today, I will formally prove the theorem.

The proof of the theorem relies on a principle that doesn’t seem to be taught very often anymore… rearranging the terms of a double sum. In this case, the double sum is

e^z e^w = \displaystyle \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{z^n}{n!} \frac{w^k}{k!}

This can be visualized in the picture below, where the x-axis represents the values of k and the y-axis represents the values of n. Each red dot symbolizes a term in the above double sum. For a fixed value of n, the values of k vary from 0 to \infty. In other words, we start with n =0 and add all the terms on the line n = 0 (i.e., the x-axis in the picture). Then we go up to n = 1 and then add all the terms on the next horizontal line. And so on.

double sum 1

I will rearrange the terms as follows: Let j = n+k. Then for a fixed value of j, the values of k will vary from 0 to j. This is perhaps best described in the picture below. The value of j, the sum of the coordinates, is constant along the diagonal lines below. The value of k then changes while moving along a diagonal line.

Even though this is a different way of adding the terms, we clearly see that all of the red circles will be hit regardless of which technique is used for adding the terms.

double sum 2

In this way, the double sum \displaystyle \sum_{n=0}^\infty \sum_{k=0}^\infty gets replaced by \displaystyle \sum_{j=0}^\infty \sum_{k=0}^j. Since n = j-k, we have

e^z e^w = \displaystyle \sum_{j=0}^{\infty} \sum_{k=0}^{j} \frac{z^{j-k}}{(j-k)!} \frac{w^k}{k!}

We now add a couple of j! terms to this expression for reasons that will become clear shortly:

e^z e^w = \displaystyle \sum_{j=0}^{\infty} \sum_{k=0}^{j} \frac{j!}{j!} \frac{1}{k! (j-k)!} w^k z^{j-k}

Since j! does not contain any ks, it can be pulled outside of the inner sum on k. We do this for the j! in the denominator:

= \displaystyle \sum_{j=0}^\infty \frac{1}{j!} \sum_{k=0}^j \frac{j!}{k!(j-k)!} w^k z^{j-k}

We recognize that \displaystyle \frac{j!}{k! (j-k)!} is a binomial coefficent:

= \displaystyle \sum_{j=0}^\infty \frac{1}{j!} \sum_{k=0}^j {j \choose k} w^k z^{j-k}

The inner sum is recognized as the formula for a binomial expansion:

= \displaystyle \sum_{j=0}^\infty \frac{1}{j!} (w+z)^j

Finally, we recognize this as the definition of e^{w+z}, using the dummy variable j instead of n. This proves that e^z e^w = e^{z+w} even if z and w are complex.

Without a doubt, this theorem was a lot of work. The good news is that, with this result, it will no longer be necessary to explicitly use the summation definition of e^z to actually compute e^z, as we’ll see tomorrow.

green line 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)

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.

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 15)

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)

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.

There’s a shorthand notation for the right-hand side (r e^{i \theta}) that, at long last, I will explain in today’s post.

Definition. If z is a complex number, then we define

e^z = \displaystyle \sum_{n=0}^{\infty} \frac{z^n}{n!}

This of course matches the Taylor expansion of e^x for real numbers x.

Theorem. If \theta is a real number, then e^{i \theta} = \cos \theta + i \sin \theta.

e^{i \theta} = \displaystyle \sum_{n=0}^{\infty} \frac{(i \theta)^n}{n!}

= \displaystyle 1 + i\theta + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \frac{(i\theta)^4}{4!} + \frac{(i\theta)^5}{5!} + \frac{(i\theta)^6}{6!} + \frac{(i\theta)^7}{7!} + \dots

= \displaystyle \left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} \dots \right) + i \left( \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} \right)

= \cos \theta + i \sin \theta,

using the Taylor expansions for cosine and sine.

This theorem explains one of the calculator’s results:

e^{i \pi} = \cos \pi + i \sin \pi = -1 + 0i = -1.

That said, you can imagine that finding something like e^{4-2i} would be next to impossible by directly plugging into the series and trying to simply the answer. The good news is that there’s an easy way to compute e^z for complex numbers z, which we develop in the next few posts.

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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 14)

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.

Definition. If z is a complex number, then we define

e^z = \displaystyle \sum_{n=0}^{\infty} \frac{z^n}{n!}

Even though this isn’t the usual way of defining the exponential function for real numbers, the good news is that one Law of Exponents remains true. (At we saw in an earlier post in this series, we can’t always assume that the usual Laws of Exponents will remain true when we permit the use of complex numbers.)

Theorem. If z and w are complex numbers, then e^z e^w = e^{z+w}.

I will formally prove this in the next post. Today, I want to talk about the idea behind the proof. Notice that

e^z e^w = \displaystyle \left( 1 + z + \frac{z^2}{2!} +\frac{z^3}{3!} + \frac{z^4}{4!} + \dots \right) \left( 1 + w + \frac{w^2}{2!} + \frac{w^3}{3!} + \frac{w^4}{4!} + \dots \right)

Let’s multiply this out (ugh!), but we’ll only worry about terms where the sum of the exponents of z and w is 4 or less. Here we go…

e^z e^w = \displaystyle 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \frac{z^4}{4!} + \dots

+ \displaystyle w + wz + \frac{wz^2}{2!} + \frac{wz^3}{3!} + \dots

+ \displaystyle \frac{w^2}{2!} + \frac{w^2 z}{2!} + \frac{w^2 z^2}{2! \times 2!} + \dots

+ \displaystyle \frac{w^3}{3!} + \frac{w^3 z}{3!} + \dots

+ \displaystyle \frac{w^4}{4!} + \dots

Next, we rearrange the terms according to the sum of the exponents. For example, the terms with z^3, w z^2, w^2 z, and w^3 are placed together because the sum of the exponents for each of these terms is 3.

e^z e^w = 1

+ z + w

\displaystyle + \frac{z^2}{2} + wz + \frac{w^2}{2}

\displaystyle + \frac{z^3}{6} + \frac{wz^2}{2} + \frac{w^2 z}{2} + \frac{w^3}{6}

\displaystyle + \frac{z^4}{24} + \frac{w z^3}{6} + \frac{w^2 z^2}{4} + \frac{w^3 z}{6} + \frac{w^4}{24} + \dots

For each line, we obtain a common denominator:

e^z e^w = 1

+ z + w

\displaystyle + \frac{z^2 + 2 z w + w^2}{2}

\displaystyle + \frac{z^3 + 3 z^2 w + 3 z w^2 + w^3}{6}

\displaystyle + \frac{z^4+ 4 z^3 w + 6 z^2 w^2 + 4 z w^3 + w^4}{24} + \dots

We recognize the familiar entries of Pascal’s triangle in the coefficients of the numerators, and so it appears that

e^z e^w = 1 + (z+w) + \displaystyle \frac{(z+w)^2}{2!} + \frac{(z+w)^3}{3!} + \frac{(z+w)^4}{4!} + \dots

If the pattern on the right-hand side holds up for exponents greater than 4, this proves that e^z e^w = e^{z+w}.

So that’s the idea of the proof. The formal proof will be presented in the next post.

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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 13)

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)

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.

There’s a shorthand notation for the right-hand side (r e^{i \theta}) that I’m about to justify.

Definition. If z is a complex number, then we define

e^z = \displaystyle \sum_{n=0}^{\infty} \frac{z^n}{n!}

This of course matches the Taylor expansion of e^x for real numbers x.

For example,

e^i = \displaystyle \sum_{n=0}^{\infty} \frac{i^n}{n!}

= \displaystyle 1 + i + \frac{i^2}{2!} + \frac{i^3}{3!} + \frac{i^4}{4!} + \frac{i^5}{5!} + \frac{i^6}{6!} + \frac{i^7}{7!} + \dots

= \displaystyle \left(1 - \frac{1}{2!} + \frac{1}{4!} - \frac{1}{6!} \dots \right) + i \left( 1 - \frac{1}{3!} + \frac{1}{5!} - \frac{1}{7!} \right)

= \cos 1 + i \sin 1,

using the Taylor expansions for cosine and sine (and remembering that this is 1 radian, not 1 degree).

This was a lot of work, and raising i to successive powers is easy! You can imagine that finding something like e^{4-2i} would be next to impossible by directly plugging into the series and trying to simply the answer.

The good news is that there’s an easy way to compute e^z for complex numbers z, which we develop in the next few posts. Eventually, this will lead to the calculation of e^{\pi i} which is demonstrated in the video below.

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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)

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)

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.

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 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.

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)

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.

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.