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1.6: Polar Equations and Complex Numbers

Difficulty Level: At Grade Created by: CK-12
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Polar Coordinates

Polar Coordinates


On example 1, you may need to stress that we are moving clockwise because the \begin{align*}\theta\end{align*}-coordinate is negative, and that otherwise we would be moving counterclockwise.

Sinusoids of One Revolution


An important thing to explain about graphing in polar coordinates is that \begin{align*}r\end{align*} is always the dependent variable and \begin{align*}\theta\end{align*} is the independent variable—so the equations we graph in polar coordinates will take forms like \begin{align*}r = \sin \theta\end{align*}, or more complex expressions based on \begin{align*}\theta\end{align*}, where \begin{align*}\theta\end{align*} is always an angle measure. The fact that \begin{align*}r\end{align*} is always written first in the ordered-pair representation of a point is a little counterintuitive, because we have previously been used to seeing the independent variable (which is usually \begin{align*}x\end{align*}) written first.

Why do we write \begin{align*}r\end{align*} first if \begin{align*}r\end{align*} is the dependent variable? One reason is that graphing points is easier, or at least more intuitive, if we look at the \begin{align*}r-\end{align*}coordinate first; we can think of ourselves as starting at the origin, moving along the \begin{align*}x-\end{align*}axis to the given \begin{align*}r-\end{align*}value, and then moving around the circle to the given \begin{align*}\theta-\end{align*}value.

(Another reason has to do with the conventions for representing complex numbers in polar form, which will be covered later in this chapter.)

Example 3 contains an interesting optical illusion: the diagonal lines passing through the graph make it look slightly warped and may prevent students from realizing that it is in fact a perfect circle.

Also worth pointing out about this example is that all of the points on the graph have in fact been traced twice over. The \begin{align*}\theta-\end{align*}values from \begin{align*}0^\circ\end{align*} to \begin{align*}180^\circ\end{align*} traced out the circle; then the \begin{align*}\theta-\end{align*}values from \begin{align*}180^\circ\end{align*} to \begin{align*}360^\circ\end{align*} produced the same set of \begin{align*}r-\end{align*}values all over again, but negative. Since \begin{align*}(-r, \theta + 180^\circ)\end{align*} always represents the same point as \begin{align*}(r, \theta)\end{align*}, the second set of \begin{align*}r-\end{align*}values correspond to the same set of points as the first set.

The important thing to point out about cardioids is that they are dimpled limaçons whose “dimple” passes directly through the pole. The depth of the dimple in a limaçon depends on the ratio \begin{align*}\frac{a}{b}\end{align*} (the smaller the ratio, the deeper the dimple), and the limaçon becomes a cardioid when the ratio equals \begin{align*}1\end{align*}. If the ratio got any smaller, the limaçon would dimple so far that it would develop a loop.

Applications, Trigonometric Tools


The constraints on example 1 should read “\begin{align*}-2\pi \le \theta \le 2 \pi\text{''}\end{align*} rather than \begin{align*}``0 \le \theta \le 2 \pi\text{''}\end{align*}.

Polar-Cartesian Transformations

Graphs of Polar Equations

Note that when we graph a basic cosine equation in polar coordinates, the domain only needs to go from to \begin{align*}\pi\end{align*} rather than \begin{align*}2 \pi\end{align*}. As noted in the previous lesson, the sine or cosine graph is traced out twice over on an interval of \begin{align*}2 \pi\;\mathrm{units}\end{align*}. In a sense, we can almost say that when we are using polar coordinates, the sine and cosine functions have a period of \begin{align*}\pi\end{align*} instead of \begin{align*}2 \pi\end{align*}.

Conic Section Transformations


An ellipse is actually the result of the intersection of a cone on one side by a plane that may or may not be parallel to the base of the cone. When the plane is parallel to the base, the ellipse is a circle.

The plane that creates a parabola cannot just be non-parallel to the base; it must be parallel to the slanted line that forms the edge of the cone.

The plane that creates a hyperbola does not actually have to be perpendicular to the base, as long as it intersects both halves of the cone.

Also, you may need to clarify that the definition of “cone” used here differs from the one students encountered in geometry: a “cone” here is really two cones lined up tip to tip, and the two cones actually extend infinitely far outward from the point where they meet.

Parabolas can theoretically be stretched horizontally as well as vertically, but stretching them horizontally is in a sense the same as shrinking (or “un-stretching”) them vertically, so it can be expressed in the same way, with a little adjusting of arbitrary constants.

The focal axis of an ellipse is also called the major axis, and its length is denoted as 2a. The perpendicular line passing through the center of the ellipse is the minor axis and its length is 2b. This will be important later.

Points to Consider

Circles centered at the origin are certainly easier to express in polar coordinates, but those that have been shifted away from the origin may be a little harder. Parabolas tend to be easier to represent in rectangular coordinates. In general, taking a familiar equation and shifting or stretching it in one direction or another tends to be easier when the equation is expressed in rectangular coordinates. (These are just a few examples; students may provide others.)

Polar curves may of course intersect, as we saw during this lesson. This question prepares students for the next lesson, where they will learn to find the intersection points of such curves.

Since two different sets (in fact, infinitely many sets) of coordinates can be used to represent the same set of points, it makes sense that two different equations could be used to represent the same polar curve. For example, \begin{align*}r = \sin \theta\end{align*} would produce the same graph as \begin{align*}r = \sin (\theta + 2\pi)\end{align*}.

One important difference between rectangular and polar representation is that polar graphs are more likely to stay within a finite viewing space. When we graph a function on a rectangular grid, if the function’s domain is unlimited, then the graph extends infinitely far to each side, so we can’t ever really draw the entire graph. Polar graphs, on the other hand, can extend infinitely far outward if the range is unlimited, but if the range is limited, then the domain can be unlimited and the graph will still be conveniently compact.

(Then again, this can be an inconvenience at times, as it makes it harder to show when we are deliberately only graphing part of a function instead of the whole thing. Since the whole thing should fit in the visible part of the graph, viewers will expect the visible part of the graph to contain the whole function unless we include a note specifying otherwise. With rectangular coordinates, we can simply narrow the graphing window to show only the part we want to graph, and viewers don’t need to be told that there’s really more to the graph than just that part.)

Systems of Polar Equations

Graph and Calculate Intersections of Polar Curves


In the solution to example 2, the notation \begin{align*}k \epsilon I\end{align*} may be unfamiliar to students; it means “\begin{align*}k\end{align*} is an element (a member) of the set of all integers,” or simply “\begin{align*}k\end{align*} is an integer.” So saying that the solution set includes \begin{align*}\left (1, \frac{5 \pi}{3}\right ) + 2 \pi k, k \epsilon I\end{align*} is simply another way of saying that when we add any integer multiple of \begin{align*}2 \pi\end{align*} to the \begin{align*}\theta-\end{align*}coordinate in the solution \begin{align*}\left (1, \frac{5 \pi}{3}\right )\end{align*}, we get another valid solution—and of course the same holds true for the solution \begin{align*}\left (1, \frac{\pi}{3}\right )\end{align*}.

The solution to Example 3 may be confusing at first—how can \begin{align*}(0,0)\end{align*} and \begin{align*}\left (0, \frac{\pi}{2}\right )\end{align*} represent the same point? Students should grasp by now that adding a multiple of \begin{align*}2 \pi\end{align*} to the \begin{align*}\theta-\end{align*}coordinate of a point yields another representation of the same point, but in this case we’ve added \begin{align*}\frac{\pi}{2}\end{align*}, which isn’t a multiple of \begin{align*}2 \pi\end{align*}—so what’s going on here?

The important thing to explain here is that the pole (as we call the origin when we are using polar coordinates) is a very special point. Normally, any given \begin{align*}r-\end{align*}coordinate designates a circle centered at the pole with radius \begin{align*}r\end{align*}, and we use different \begin{align*}\theta-\end{align*}coordinates to pick out specific points on the circle. But a circle with radius is just a single point—the pole itself—so no matter what \begin{align*}\theta-\end{align*}coordinate we choose, we always end up at that same point. \begin{align*}(0,0)\end{align*} represents the same point as \begin{align*}(0,3), \left (0, \frac{\pi}{2}\right ), (0, 4\pi),\end{align*} or any ordered pair whatsoever that has as the \begin{align*}r-\end{align*}coordinate.

Points to Consider

A very simple example of two polar curves that do not intersect is the pair \begin{align*}r = 1\end{align*} and \begin{align*}r = 2\end{align*}. And we have seen again in this lesson how the same point can be expressed in more than one way in polar coordinates; in the next lesson we will see how the same curve can too.

Equivalent Polar Curves

It is important that students do not get the mistaken idea that expressions with equivalent graphs are necessarily equivalent expressions. In example 1b, the two equations graphed are indeed equivalent, as they are both simply different ways of expressing \begin{align*}r = 5\end{align*}. But in example 1a, although the two equations trace out the same graph, they do not actually have the same \begin{align*}r-\end{align*}value for any given \begin{align*}\theta-\end{align*}value, and so are not equivalent equations. If they were equivalent, plugging the same \begin{align*}\theta-\end{align*}value into both of them would always yield the same \begin{align*}r-\end{align*}value.

This is especially confusing because it only happens in polar coordinates, where the \begin{align*}\theta\end{align*}-values overlap and repeat themselves. Here’s an analogy that may help you to explain it: Suppose you ride the same bus to work or school every day, and suppose the bus maintains a very strict schedule, so it always reaches the same stop on its route at exactly the same time each day. (Of course no real bus could manage this, but let’s assume it does in order to simplify the problem.) Now suppose you draw a graph each day representing the route the bus travels, and when you compare two consecutive graphs, you see that they look exactly the same—if you plotted them on the same axes, they would look like just one graph. But does this mean they are describing the exact same trip? No—they represent two different trips taken on two different days, and when you plot them on the same axis you are simply leaving out the “two different days” part. Really, the time-values of the second graph are the time-values of the first graph “plus \begin{align*}24\;\mathrm{hours}\end{align*},” and if you choose to plot them on the same graph to save space, you must still remember that the times on the two graphs aren’t “really” the same.

And that’s what happens when we use polar coordinates—we sometimes end up graphing the \begin{align*}\theta-\end{align*}values for two or more different \begin{align*}r-\end{align*}values in what looks like the same spot, but we must remember that just because they are sharing space, that doesn’t mean they are really the same coordinates. Even when a whole graph looks the same as another, sometimes it simply consists of a different set of values that happen to be graphed in the same places.

Imaginary and Complex Numbers



Here’s another example of why the rule \begin{align*}\sqrt{ab} = \sqrt{a} \sqrt{b}\end{align*} only applies if \begin{align*}a\end{align*} and \begin{align*}b\end{align*} are not both negative: Without that exception, we could apply the rule to \begin{align*}\sqrt{36}\end{align*} and express it as \begin{align*}\sqrt{-4} \sqrt{-9}\end{align*}, which would equal \begin{align*}2i\;\mathrm{times}\ 3i\end{align*}, or \begin{align*}-6\end{align*}. Technically, \begin{align*}(-6)^2\end{align*} is of course \begin{align*}36\end{align*}, but officially -6 is not the square root of 36, so that answer would be incorrect.

The last line of the solution to example 2b should have a \begin{align*}5\end{align*}, rather than a \begin{align*}3\end{align*}, under the radical sign.

Points to Consider

Students needn’t know the answers to these questions; they are simply a preview of the next section.

Standard Form of Complex Numbers (a + bi)


Students may be a little confused by the statement that a and b are both real numbers in the standard form \begin{align*}a + bi\end{align*}. Clarify if necessary that the imaginary part of a complex number is \begin{align*}bi\end{align*}, not just \begin{align*}b\end{align*}; \begin{align*}bi\end{align*} is a pure imaginary number because it is a real number multiplied by \begin{align*}i\end{align*}.

Students may not quite see why the answer to example 1c is expressed as it is. The answer is indeed in standard rectangular form, with \begin{align*}a = 3\sqrt{2}\end{align*} and \begin{align*}b = -2\sqrt{2}\end{align*}, but we traditionally put the \begin{align*}i\end{align*} in front of the radical sign so that it doesn’t look like it is included under the radical sign, and that makes it harder to see that the whole expression has the form \begin{align*}a + bi\end{align*}.

After reviewing example 3, you might also want to challenge students to find the conjugate of a real number, like \begin{align*}5\end{align*}. (Answer: \begin{align*}5\end{align*} is really \begin{align*}5 + 0i\end{align*}, so its conjugate is \begin{align*}5 - 0i\end{align*}, or simply \begin{align*}5\end{align*} again. In other words, the conjugate of any real number is simply itself.)

Points to Consider

We will see in the next lesson what operations can be performed on complex numbers and with what results.

Complex Number Plane

If you’ve covered vectors recently, you might point out here that the absolute value of the complex number \begin{align*}a + bi\end{align*} is the same as the magnitude of the vector represented by the point \begin{align*}(a,b)\end{align*}.

You may need to skip over the problem about two students walking home, since the original formulation of the problem is missing from the lesson.

Operations on Complex Numbers

Quadratic Formula

Points to Consider

When the roots of an equation are complex, we know that the graph of the equation does not intersect the \begin{align*}x-\end{align*}axis. Conversely, when the graph does not intersect the \begin{align*}x-\end{align*}axis, we know the roots are complex, and when it does, we know there are either two real roots or one real root repeated twice.

Sums and Differences of Complex Numbers

Points to Consider

We will see in the next lesson how complex numbers can be expressed in polar form.

Products and Quotients of Complex Numbers (conjugates)


There is a typographical error in the formula for multiplying complex numbers: the term that reads \begin{align*}(ad - bd)\end{align*} should read \begin{align*}(ac - bd)\end{align*}.

Students may not quite see where the \begin{align*}``-bd\text{''}\end{align*} term comes from. Explain that multiplying \begin{align*}bi\end{align*} and \begin{align*}di\end{align*} yields \begin{align*}bdi^2\end{align*}, and \begin{align*}i^2\end{align*} is simply \begin{align*}-1\end{align*}, leaving us with \begin{align*}-bd\end{align*}.

The procedure for dividing complex numbers may make more sense if you remind students that \begin{align*}i\end{align*} is equal to \begin{align*}\sqrt{-1}\end{align*}. When we express the quotient of two complex numbers as a fraction, substituting \begin{align*}\sqrt{-1}\end{align*} for \begin{align*}i\end{align*} shows that this fraction essentially has a radical in the denominator which we must rationalize. Multiplying the numerator and denominator by the conjugate of the denominator, then, is clearly the way to get the imaginary part out of the denominator; and once the denominator is a real number, we can divide both parts of the numerator by that real number and thus express the answer in standard form.

Points to Consider

Once again, we will see in the next lesson that the answer to both of these questions is “yes.”

Applications, Trigonometric Tools

Operations on Complex Numbers

Example 2 provides an excellent illustration of the relationship between operations on complex numbers and operations on vectors. You may want to point out to students that when they solve a problem like this by working with the real and imaginary parts of complex numbers separately, they are really just resolving vectors into horizontal and vertical components and working with each component separately. But instead of using \begin{align*}\hat i\end{align*} and \begin{align*}\hat j\end{align*} to represent those components, they are using \begin{align*}1\end{align*} and \begin{align*}i\end{align*}, because the horizontal component is a multiple of the unit vector in the real direction or “\begin{align*}1-\end{align*}direction” and the vertical component is a multiple of the unit vector in the imaginary or “\begin{align*}i-\end{align*}direction.”

Trigonometric Form of Complex Numbers

Trigonometric Form of Complex Numbers: Relationships among \begin{align*}x, y, r,\end{align*} and \begin{align*}\theta\end{align*}

In case students don’t immediately see why \begin{align*}x = r \cos \theta\end{align*} and \begin{align*}y = r \sin \theta\end{align*}, you can show them fairly easily on the diagram that \begin{align*}\sin \theta = \frac{y}{r}\end{align*}, and then solving for \begin{align*}y\end{align*} yields \begin{align*}y = r \sin \theta\end{align*}. Similar reasoning, of course, holds for \begin{align*}x\end{align*}.

The Trigonometric or Polar Form of a Complex Number \begin{align*}(r cis \theta)\end{align*}

The term \begin{align*}``\mathrm{cis} \theta\text{''}\end{align*} is easier to remember if you point out that it is somewhat like an acronym, derived from \begin{align*}``\cos \theta + i \sin \theta.\text{''}\end{align*}

The term “argument” is also worth explaining, as it often appears elsewhere in mathematics. Generally it refers to the “input” of a given function, so for example, in the expression \begin{align*}``\cos \theta,\text{''}\end{align*} \begin{align*}\theta\end{align*} is called the argument of the cosine function. In the polar form of a complex number, of course, \begin{align*}\theta\end{align*} appears as the argument of both the sine and cosine functions, so it makes a kind of sense to call it the “argument” of the complex number as a whole.

Thinking of r as the “absolute value” of a complex number may be counterintuitive for students, but it is really just an extension of the idea of absolute value of real numbers. The absolute value of a real number is its distance from ; the absolute value of a complex number is its distance from the point \begin{align*}(0,0)\end{align*}. And in the complex plane, the distance of a real number from becomes the same thing as its distance from \begin{align*}(0,0)\end{align*}.

Trigonometric Form of Complex Numbers: Steps for Conversion


The very first table in this section is the most useful for students to know, and it is most helpful for them to think of the left half and the right half separately. Emphasize that if they know the polar coordinates \begin{align*}r\end{align*} and \begin{align*}\theta\end{align*}, they should use the two equations on the left to find the rectangular coordinates \begin{align*}x\end{align*} and \begin{align*}y\end{align*}, whereas if they know \begin{align*}x\end{align*} and \begin{align*}y\end{align*} they should use the two equations on the right to find \begin{align*}r\end{align*} and \begin{align*}\theta\end{align*}. (They could use the equations on the right to get \begin{align*}x\end{align*} and \begin{align*}y\end{align*} from \begin{align*}r\end{align*} and \begin{align*}\theta\end{align*}, or use the equations on the left to get \begin{align*}r\end{align*} and \begin{align*}\theta\end{align*} from \begin{align*}x\end{align*} and \begin{align*}y\end{align*}, but that would be a much messier process.)

The last step of Example 3 is important: finding the inverse tangent just tells us the reference angle for \begin{align*}\theta,\end{align*} not \begin{align*}\theta\end{align*} itself. We must then apply our knowledge of what quadrant the complex number is in to figure out what angle \begin{align*}\theta\end{align*} really is. We can find out what quadrant the number is in by graphing the rectangular coordinates we started out with, or by simply noting the signs of those coordinates: \begin{align*}x\end{align*} is only positive in the first two quadrants, and \begin{align*}y\end{align*} is only positive in the first and fourth.

Points to Consider

In the next lesson, we will see how to perform basic operations such as multiplication and division on complex numbers in polar form.

Product and Quotient Theorems

Product Theorem

The first line of the derivation here uses the FOIL method for multiplying binomials; in the second line, we group together the terms with \begin{align*}i\end{align*} in them and factor out the \begin{align*}i\end{align*}; and in the third line, we apply the angle sum rules for sine and cosine in reverse.

The Product Theorem may be easier for students to remember if summarized in words: “To multiply complex numbers in polar form, multiply the \begin{align*}r-\end{align*}coordinates and add the \begin{align*}\theta-\end{align*}coordinates.”

Quotient Theorem

Similarly, the Quotient Theorem can be summarized: “To divide complex numbers in polar form, divide the \begin{align*}r-\end{align*}coordinates and subtract the \begin{align*}\theta-\end{align*}coordinates.” The derivation follows much the same procedure as the one for the product theorem; note that the denominator of the final fraction equals \begin{align*}1\end{align*}, so we can cancel it out.

Incidentally, some students may notice that we haven’t discussed how to add and subtract complex numbers in polar form. It turns out that there is no handy formula for doing so; the only good way to add and subtract complex numbers is to convert them to rectangular form first.

Using the Quotient and Product Theorem


Students may be confused by example 3, or may be tempted to do the division on the numbers as given, in rectangular form. Doing it as the book suggests, however, will give them practice in converting from rectangular to polar form as well as in performing division on numbers in polar form. Meanwhile, the other three examples will give them practice in working with complex numbers expressed in polar form in more than one way.

For extra practice, you might have them find both the product and the quotient of the two numbers in each example, instead of just the product or the quotient.

Points to Consider

So far, we’ve actually just barely touched on squares and square roots of complex numbers, and we still don’t know how to find them for most numbers. The next lesson will cover these as well as other powers and roots of complex numbers.

Applications and Trigonometric Tools: Real-Life Problem

You may need to skip these problems, as they contain terms that students are not likely to know .

Powers and Roots of Complex Numbers

De Moivre’s Theorem

A more intuitive way to express De Moivre’s Theorem is “To raise a complex number to the \begin{align*}n^{\mathrm{th}}\end{align*} power, raise the \begin{align*}r-\end{align*}coordinate to the \begin{align*}n^{\mathrm{th}}\end{align*} power and multiply the \begin{align*}\theta-\end{align*}coordinate by \begin{align*}n\end{align*}.”

In example 2, the expression \begin{align*}\left (\frac{-1}{2} + \frac{\sqrt{3}} {2}\right )\end{align*} should read \begin{align*}\left (\frac{-1}{2} + \frac{\sqrt{3}} {2} i \right )\end{align*}.

nth Root Theorem

Here’s a much more intuitive way to explain the \begin{align*}n^{\mathrm{th}}\end{align*} Root Theorem:

Every complex number has exactly \begin{align*}n \ n^{\mathrm{th}}\end{align*} roots, which are evenly spaced around a circle in the complex plane. If the original number has coordinates \begin{align*}(r, \theta)\end{align*}, then the first of the \begin{align*}n^{\mathrm{th}}\end{align*} roots (which, incidentally, is known as the principal root) has coordinates \begin{align*}\left (\sqrt[n]{r}, \frac{\theta}{n}\right )\end{align*}. The rest of the \begin{align*}n^{\mathrm{th}}\end{align*} roots all have the same \begin{align*}r-\end{align*}coordinate, and their \begin{align*}\theta-\end{align*}coordinates are each \begin{align*}\frac{\theta}{n}\end{align*} plus some multiple of \begin{align*}\frac{2 \pi}{n}\end{align*}; in other words, each of them is \begin{align*}\frac{1}{n}\end{align*} of the way around the circle from the one before it.

For example, the fourth roots of \begin{align*}(16, 60^\circ)\end{align*} are \begin{align*}(2, 15^\circ), (2, 105^\circ), (2, 195^\circ),\end{align*} and \begin{align*}(2, 285^\circ)\end{align*}. Plotting these points shows that they are evenly spaced around a circle of radius \begin{align*}2\end{align*}, and a little thought will show why. Raising \begin{align*}2\end{align*} to the fourth power of course gives us \begin{align*}16\end{align*}, and multiplying an angle of \begin{align*}15^\circ\end{align*} by \begin{align*}4\end{align*} gives us \begin{align*}60^\circ\end{align*} —but multiplying an angle of \begin{align*}15^\circ\end{align*}-plus-some-multiple-of-\begin{align*}90^\circ\end{align*} by \begin{align*}4\end{align*} gives us \begin{align*}60^\circ\end{align*}-plus-some-multiple-of-\begin{align*}360^\circ\end{align*}, which is equivalent to \begin{align*}60^\circ\end{align*}. So that’s why, if \begin{align*}(2, 15^\circ)\end{align*} is one fourth root of \begin{align*}(16, 60^\circ)\end{align*}, all the other fourth roots are of the form \begin{align*}(2, 15^\circ + k*90^\circ)\end{align*} for some integer \begin{align*}k\end{align*}. (If they were fifth roots, they’d be \begin{align*}\frac{360^\circ}{5}\end{align*} or \begin{align*}72^\circ\end{align*} apart; if they were sixth roots they’d be \begin{align*}60^\circ\end{align*} apart, and so on.)

Incidentally, we could just keep going, adding \begin{align*}90^\circ\end{align*} to each previous \begin{align*}\theta-\end{align*}coordinate to get yet another one. But of course, after we’ve added \begin{align*}90^\circ\end{align*} three times to the first solution to get three more solutions, adding \begin{align*}90^\circ\end{align*} one more time would just give us the first solution plus \begin{align*}360^\circ\end{align*}, or in other words the first solution all over again, and then we’d start cycling through all the solutions over again. So the first four solutions (the first \begin{align*}n\end{align*} solutions, if we’re finding \begin{align*}n^{\mathrm{th}}\end{align*} roots) are the only unique ones, and we can stop after finding them.

Solve Equations

It should become clear here that we can use polar coordinates to determine the \begin{align*}n^{\mathrm{th}}\end{align*} roots of a pure real number or pure imaginary number just as easily as a complex number. However, the roots are usually complex and tend to be somewhat messy to express in rectangular coordinates. For example, in the problem shown in the text, we can see that the roots can be expressed precisely in polar coordinates, and that we can use just one expression to summarize them all, whereas in rectangular coordinates we must list all the roots separately and can only express them in decimal approximations.

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