Calculators and complex numbers (Part 23)

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.

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 22)

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

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

At long last, we are now in position to explain the last surprising results from the calculator video below.

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

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

Naturally, this definition makes sense if $z$ and $w$ are real numbers.

For example, let’s consider the computation of $i^i$ . For the base of $i$ , we note that

$\log i = \log e^{\pi i/ 2} = \displaystyle \frac{\pi i}{2}$ .

Therefore,

$i^i = e^{i \log i} = e^{i \pi i/2} = e^{-\pi/2}$ ,

which is (surprisingly) a real number.

As a second example, let’s compute $(-8)^i$ . To begin,

$\log(-8) = \log \left( 8 e^{\pi i} \right) = \ln 8 + \pi i$ .

Therefore,

$(-8)^i = e^{i \log(-8)}$

$= e^{i (\ln 8 + \pi i)}$

$= e^{-\pi + i \ln 8}$

$= e^{-\pi} (\cos [\ln 8] + i \sin [ \ln 8 ] )$

$= e^{-\pi} \cos (\ln 8) + i e^{-\pi} \sin (\ln 8)$

In other words, a problem like this is a Precalculus teacher’s dream come true, as it contains $e, \ln, \pi, \cos, \sin$ , and $i$ in a single problem.

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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 21)

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

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

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

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 $r e^{i \theta}$ . However, this complex logarithm doesn’t always work the way you’d think it work. For example,

$\log \left(e^{2 \pi i} \right) = \log (\cos 2\pi + i \sin 2\pi) = \log 1 = \ln 1 = 0 \ne 2\pi i$ .

This is analogous to another situation when an inverse function is defined using a restricted domain, like

$\sqrt{ (-3)^2 } = \sqrt{9} = 3 \ne -3$

$\sin^{-1} (\sin \pi) = \sin^{-1} 0 = 0 \ne \pi$ .

The Laws of Logarithms also may not work when nonpositive numbers are used. For example,

$\log \left[ (-1) \cdot (-1) \right] = \log 1 = 0$ ,

but

$\log(-1) + \log(-1) = \log \left( e^{\pi i} \right) + \log \left( e^{\pi i} \right) = \pi i + \pi i = 2\pi i$ .

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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 20)

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

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

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)

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

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

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)

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

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

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)

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.

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.

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 $k$ s, 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)

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 15)

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, 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)

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

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

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.