# 2.6: Polar Equations and Complex Numbers

**At Grade**Created by: CK-12

## Polar Coordinates

*Polar Coordinates*

**In-Text Examples**

2) You may want to stress that if \begin{align*}(4, 120^\circ)\end{align*} is one way of representing the given point, \begin{align*}(-4, -120^\circ)\end{align*} is *not* another way. In general, if we change the sign of the \begin{align*}r-\end{align*}coordinate, we cannot simply change the sign of the \begin{align*}\theta-\end{align*}coordinate to keep the point the same.

(Here’s part of the reason why: When we change the sign of the \begin{align*}r-\end{align*}coordinate, we’re actually rotating the point \begin{align*}180^\circ\end{align*} about the origin—or reflecting it across both the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}axes, which is the same thing. When we change the sign of the \begin{align*}\theta-\end{align*}coordinate, we’re reflecting the point across the \begin{align*}x-\end{align*}axis. So if we change the signs of both coordinates, those two transformations combined add up to one reflection across the \begin{align*}y-\end{align*}axis. Explaining this is optional, though; the important part is that changing the signs of both coordinates doesn’t get us back to the same point.)

**Review Questions**

2) Students’ answers should include the original coordinates \begin{align*}\left (-4, \frac{\pi}{4}\right )\end{align*} rather than the third pair of coordinates given in the answer key.

*Sinusoids of One Revolution*

**In-Text Examples**

5) Students may have trouble graphing looped limaçons like this if they don’t include enough values in their table. Also, it may be hard for them to keep track of which order the points should be connected in, especially when the \begin{align*}r-\end{align*}values become negative.

**Review Questions**

1) All the curves here are limaçons, but that answer is not sufficient.

2) The rose will have \begin{align*}n\end{align*} petals if \begin{align*}n\end{align*} is odd and \begin{align*}2n\end{align*} petals if \begin{align*}n\end{align*} is even, but that may not be apparent just from the two examples given. Students shouldn’t get hung up on trying to find a more specific relationship than the one given in the answer key.

*Applications, Trigonometric Tools*

**In-Text Examples**

2) Students may think that the total distance \begin{align*}\theta\end{align*} they must plug into the area equation is \begin{align*}\frac{\pi}{3}\end{align*} rather than \begin{align*}\frac{2 \pi}{3}\end{align*}, because of the \begin{align*}\frac{\pi}{3}\end{align*} coordinate used earlier. The graph should help them avoid this mistake, though.

3) Any three roses will do, not just the three shown. Creating a quilt can’t easily be done on just one set of axes, though; it will need to be done by copying one graph several times onto a sheet of paper.

## Polar-Cartesian Transformations

*Polar to Rectangular*

**Review Questions**

1) It’s common to get the equations \begin{align*}x = r \cos \theta\end{align*} and \begin{align*}y = r \sin \theta\end{align*} backwards when solving a problem, and think that \begin{align*}y = r \cos \theta\end{align*} and \begin{align*}x = r \sin \theta\end{align*}; also, it’s common to think of \begin{align*}\tan \theta\end{align*} as being equal to \begin{align*}\frac{x}{y}\end{align*} instead of \begin{align*}\frac{y}{x}\end{align*}. Watch for these errors on all problems of this type.

Also, if students use calculators on this problem they may forget to switch modes on part B.

*Rectangular to Polar*

**In-Text Examples**

1) You may need to stress the part about adding \begin{align*}\pi\end{align*} to the arctangent when \begin{align*}x\end{align*} is less than \begin{align*}0\end{align*}, so that students remember it when solving future problems.

**Review Questions**

1) Students working in degrees instead of radians will get \begin{align*}A(5.39, 111.73^\circ)\end{align*} and \begin{align*}B(6.40, -38.39^\circ)\end{align*} as their answers.

*Conic Section Transformations*

**In-Text Examples**

1) Mixing up the \begin{align*}x\text{'}\mathrm{s}\end{align*} and \begin{align*}y\text{'}\mathrm{s}\end{align*} is a common error when dealing with conics. When working with parabolas in particular, students may default to treating every one (if they don’t sketch it out carefully first) as if it opened up or down instead of left or right.

5) You may want to point out that plugging in a negative value for \begin{align*}a\end{align*} doesn’t yield a circle with a negative radius (what would that even look like?), but rather yields a circle of radius \begin{align*}a\end{align*} that’s on the “other” side of the origin—left instead of right for a cosine equation, and down instead of up for a sine equation. (This makes sense if you picture what happens to a graph like this as a decreases toward zero and finally passes zero—the circle shrinks closer and closer to the origin until it passes through the origin and comes out the other side. It also makes sense if you remember that multiplying \begin{align*}r-\end{align*}values by \begin{align*}-1\end{align*} is the same as rotating the graph \begin{align*}180^\circ\end{align*}.)

**Review Questions**

1) Students may try graphing the equation to prove it is a parabola, but this is more time-consuming and prone to error than the method described in the answer key.

2) Again, watch for students mixing up \begin{align*}x\end{align*} and \begin{align*}y\end{align*}, or \begin{align*}a\end{align*} and \begin{align*}b\end{align*}, or \begin{align*}h\end{align*} and \begin{align*}k\end{align*}.

*Applications, Technological Tools*

**Rectangular Form or Polar Form**

3) You may need to explain that the sun is at one focus of the ellipse, so the perihelion is the distance from \begin{align*}F\end{align*} to \begin{align*}P\end{align*} and the aphelion is the length of the major axis minus the perihelion.

## Systems of Polar Equations

*Graph and Calculate Intersections of Polar Curves*

**In-Text Examples**

3) Students may be very confused by the idea that \begin{align*}(0,0)\end{align*} and \begin{align*}\left (0, \frac{\pi}{2}\right )\end{align*} represent the same point—it makes sense that adding \begin{align*}2 \pi\end{align*} to the \begin{align*}\theta-\end{align*}coordinate of a point yields another name for the same point, but it doesn’t make sense that adding \begin{align*}\frac{\pi}{2}\end{align*} would do so. In fact, the only reason it works in this case is that the \begin{align*}r-\end{align*}coordinate of the point is \begin{align*}0\end{align*}. If we rotate through any angle whatsoever, but then go \begin{align*}0\;\mathrm{units}\end{align*} from the origin, we end up at the origin—so any pair of coordinates where the \begin{align*}r-\end{align*}coordinate is \begin{align*}0\end{align*} is just one more way of naming the point \begin{align*}(0, 0)\end{align*}, no matter what the \begin{align*}\theta-\end{align*}coordinate is.

4) It may seem a little strange that the polar and rectangular coordinates for the two points of intersection are exactly the same. However, this is generally true for any point whose \begin{align*}\theta-\end{align*}coordinate is \begin{align*}0\end{align*}: the \begin{align*}y-\end{align*}coordinate in rectangular form will also be \begin{align*}0\end{align*}, and the \begin{align*}x-\end{align*}coordinate will be the same as the \begin{align*}r-\end{align*}coordinate. Doing the conversion algebraically will confirm this, and it can be handy to know.

**Review Questions**

1) The graph of these equations is deceptive; it certainly looks as though they intersect three times rather than just one. But tracing the graph of \begin{align*}r = 3\ \sin \theta\end{align*} will show that at two of the apparent intersection points, the \begin{align*}r-\end{align*}value is actually negative and so is not the same as the \begin{align*}r-\end{align*}value of \begin{align*}\sin 3 \theta\end{align*} at the same point. Solving the system algebraically will also show that there is really only one solution.

*Equivalent Polar Curves*

**In-Text Examples**

1) Part b should read \begin{align*}5 \cos (-90)\end{align*} rather than \begin{align*}2\end{align*}; also, the number \begin{align*}90\end{align*} may make it seem like we are working in degrees here, but we’re actually still working in radians. (The equations are basically equivalent to \begin{align*}r = 2.24\end{align*} and \begin{align*}r = -2.24\end{align*} respectively. If we were working in degrees, those same equations would be equivalent to \begin{align*}r = 5\end{align*} and \begin{align*}r = -5\end{align*}.)

In general, the graph of \begin{align*}r = a\end{align*} is equivalent to the graph of \begin{align*}r = -a\end{align*}.

**Review Questions**

2) There’s a small chance students will get mixed up here and substitute \begin{align*}\theta\end{align*} for \begin{align*}\pi\end{align*} in their calculations, because they are used to \begin{align*}r\end{align*} being expressed in terms of \begin{align*}\theta\end{align*}. In fact, the equations are simpler than they look; since \begin{align*}\frac{\pi}{3}\end{align*} is a constant, the whole right-hand side of each equation is a constant (it works out to simply \begin{align*}5.5\end{align*}), so the graph is just a circle centered at the origin.

*Applications, Technological Tools*

**In-Text Examples**

You might want to point out in the example here that we are concerned with both the “real” points of intersection and the “apparent” ones, since this is a real-world problem where we are concerned with the shape of the graph and not just the actual number values of the points.

## Imaginary and Complex Numbers

*Recognize*

**In-Text Examples**

A very common error when simplifying square roots is to pull numbers or expressions out from under the radical sign without actually taking their square roots; this tends to happen when there is more than one expression under the radical sign to deal with. For instance, in example 1a, students might express the answer as \begin{align*}16i\end{align*} rather than \begin{align*}4i\end{align*}; or in example 2c, they might express the answer as \begin{align*}ix\end{align*} rather than \begin{align*}i \sqrt{x}\end{align*}.

**Review Questions**

1) On parts c and d, students might make the error they were warned against in the text and come up with positive \begin{align*}15\end{align*} and \begin{align*}35\end{align*} as answers.

*Standard Form of Complex Numbers \begin{align*}(a + bi)\end{align*}*

**In-Text Examples**

2) Students are likely not to fully grasp that they can treat the real terms and the imaginary terms completely separately, and that this equation therefore can really be thought of as two separate equations, one which can be solved for \begin{align*}x\end{align*} and the other for \begin{align*}y\end{align*}. They may also forget momentarily that \begin{align*}i\end{align*} is not a variable, and start vaguely trying to solve for it as well as for \begin{align*}x\end{align*} and \begin{align*}y\end{align*}, or may just get the idea that there is one equation here with three variables and hence not enough information to find a solution.

**Additional Problems**

1) Find the conjugate of each complex number:

a) \begin{align*}3 + 0i\end{align*}

b) \begin{align*}10\end{align*}

**Answers to Additional Problems**

1) a) \begin{align*}3 - 0i\end{align*} or simply \begin{align*}3\end{align*}

b) \begin{align*}10 - 0i\end{align*} or simply \begin{align*}10\end{align*}

*The Set of Complex Numbers*

**Additional Problems**

1) What sets does each number belong to?

a) \begin{align*}\sqrt{25}\end{align*}

b) \begin{align*}\frac{\pi}{3}\end{align*}

c) \begin{align*}\sin \frac{\pi}{3}\end{align*}

d) \begin{align*}\cos \frac{\pi}{4}\end{align*}

**Answers to Additional Problems**

1) a) complex, real, rational, integer, whole number, natural number

b) complex, real

c) complex, real, rational, fraction

d) complex, real

*Complex Number Plane*

**Review Questions**

1) The absolute values given are for points \begin{align*}A\end{align*} and \begin{align*}E\end{align*}; students may have chosen other points instead, so those answers are also acceptable.

## Operations on Complex Numbers

*Quadratic Formula*

**In-Text Examples**

Sign errors are very likely to occur when working with the quadratic formula.

**Review Questions**

1) Students may forget that one side of the equation must equal zero before they can apply the quadratic formula, so they may try to simply read off the coefficients from the left-hand side of the equation.

**Additional Problems**

1) a) What does the discriminant of \begin{align*}-x^2 + 6x - 9\end{align*} tell you about its root(s)?

b) Calculate the root(s).

c) Graph the equation.

**Answers to Additional Problems**

1) a) The discriminant is zero, so the function has one repeated real root.

b) The one root is \begin{align*}3\end{align*}.

c)

*Sums and Differences of Complex Numbers*

**Review Questions**

1) The answers students get from adding the vectors graphically may not be as precise as the answers they get from adding the numbers algebraically, so they may not quite match each other. This is fine as long as they are reasonably close.

*Products and Quotients of Complex Numbers (conjugates)*

**In-Text Examples**

1) This is another likely place for sign errors: when multiplying complex numbers and simplifying the answer, students are quite likely to convert \begin{align*}i^2\end{align*} to \begin{align*}1\end{align*} instead of \begin{align*}1\end{align*}, and so simply drop the \begin{align*}i^2\end{align*} altogether without changing the sign of its coefficient.

## Trigonometric Form of Complex Numbers

*Trigonometric Form of Complex Numbers: Steps for Conversion*

**In-Text Examples**

3) After learning to work with square roots of negative numbers, students may get a bit confused when finding \begin{align*}r\end{align*} based on negative values of \begin{align*}x\end{align*} and \begin{align*}y\end{align*}; they might forget that squaring those negative values should yield positive numbers, and that they should not be trying to take the square roots of any negative numbers to find \begin{align*}r\end{align*}.

**Review Questions**

2) Graphing the number on a rectangular coordinate graph is one option students may take, although that requires expressing it in standard form first and is slightly harder than graphing it in polar form.

**Additional Problems**

1) Express the sum of \begin{align*}3 + 6i\end{align*} and \begin{align*}9 - 2i\end{align*} in polar form.

**Answers to Additional Problems**

1) The sum of the two numbers is \begin{align*}12 + 4i\end{align*}, so in rectangular form \begin{align*}x = 12\end{align*} and \begin{align*}y = 4\end{align*}.

\begin{align*}r & = \sqrt{x^2 + y^2}\\ & = \sqrt{12^2 + 4^2}\\ & = \sqrt{144 + 16}\\ & = \sqrt{160}\\ & = 4 \sqrt{10}\end{align*}

\begin{align*}\tan \theta & = \frac{y}{x}\\ & = \frac{4}{12}\\ & = \frac{1}{3}\\ \theta & = \tan^{-1} \frac{1}{3}\\ & \approx 18.43^\circ\end{align*}

So in polar form the number is \begin{align*}4\sqrt{10}\end{align*} cis \begin{align*}18.43^\circ\end{align*}.

## Product and Quotient Theorems

*Using the Quotient and Product Theorem*

**In-Text Examples**

1) When applying the product rule, probably the easiest error for students to make is to get confused about whether to add the \begin{align*}r-\end{align*}values and multiply the \begin{align*}\theta-\end{align*}values or vice versa. Students may also try to add them both or multiply them both.

(Similar errors will occur when applying the quotient rule.)

3) Students may get confused when the numbers to be multiplied or divided are presented in rectangular rather than polar form; they may try to treat the real parts like r-coordinates and the imaginary parts like \begin{align*}\theta-\end{align*}coordinates, rather than converting to polar form first so they can do the calculations properly.

**Additional Problems**

1) Find the quotient: \begin{align*}(7 + 2i \sqrt{2}) \div (\sqrt{3} - 4i)\end{align*}. Express your answer in rectangular form rounded to two decimal places.

**Answers to Additional Problems**

1) First, convert to polar form:

\begin{align*}r_1 & = \sqrt{x^2 + y^2} && r_2 = \sqrt{x^2 + y^2}\\ & = \sqrt{7^2 + (2\sqrt{2})^2} && = \sqrt{\sqrt{3}^2 + (-4)^2}\\ & = \sqrt{4 + 8} &&= \sqrt{3 + 16}\\ & = \sqrt{57} &&= \sqrt{19}\end{align*}

\begin{align*}\frac{r_1}{r_2} = \frac{\sqrt{57}} {\sqrt{19}} = \frac{\sqrt{19.3}} {\sqrt{19}} = \frac{\sqrt{19}\sqrt{3}} {\sqrt{19}} = \sqrt{3}\end{align*}

\begin{align*}\theta_1 & = \tan^{-1} \frac{y}{x} && \theta_0 = \tan^{-1} \frac{y}{x}\\ & = \tan^{-1} \frac{2\sqrt{2}} {7} && = \tan^{-1} \frac{4}{\sqrt{3}}\\ & \approx 22.002^\circ && \approx 66.587^\circ\end{align*}

\begin{align*}\theta_1 - \theta_2 \approx -44.585^\circ\end{align*}

So the quotient in polar form is \begin{align*}\sqrt{3}\ \mathrm{cis} -44.585^\circ.\end{align*}

Converting back to rectangular form:

\begin{align*}x & = \sqrt{3} \cos (-44.585^\circ) && y = \sqrt{3} \sin (-44.585^\circ)\\ & \approx 1.23 && \approx -1.22\end{align*}

So the final answer is \begin{align*}1.23 -1.22i\end{align*}.

## Powers and Roots of Complex Numbers

*De Moivre’s Theorem*

**In-Text Examples**

1) As with the product theorem, students may get confused about which coordinate to multiply by \begin{align*}n\end{align*} and which one to raise to the \begin{align*}n^{\mathrm{th}}\end{align*} power when applying De Moivre’s Theorem. Also, even after writing down \begin{align*}``\cos n \theta\text{''}\end{align*} and \begin{align*}``\sin n \theta,\text{''}\end{align*} they may still try to find \begin{align*}n \ \cos \theta\end{align*} and \begin{align*}n\ \sin \theta\end{align*} instead.

**Review Questions**

1) As with the product theorem once more, students may forget to convert to polar form before applying De Moivre’s Theorem…

2) …or may forget to apply De Moivre’s Theorem before converting to rectangular form.

*nth Root Theorem*

**Additional Problems**

1) Find all the fourth roots of \begin{align*}81\end{align*}.

2) Find all the sixth roots of \begin{align*}64i\end{align*}.

**Answers to Additional Problems**

1) \begin{align*}3, 3i, -3, -3i\end{align*}

2) The principal root is \begin{align*}2\;\mathrm{cis} \frac{\pi}{6}\end{align*}, or \begin{align*}2 \left (\frac{\sqrt{3}} {2} + \frac{i}{2}\right )\end{align*}; the five other roots are \begin{align*}2\;\mathrm{cis} \frac{\pi}{2}, 2\;\mathrm{cis} \frac{5 \pi}{6}, 2\;\mathrm{cis} \frac{7 \pi}{6}, 2\;\mathrm{cis} \frac{3 \pi}{2},\end{align*} and \begin{align*}2\;\mathrm{cis} \frac{11 \pi}{6}\end{align*}.

*Solve Equations*

**Review Questions**

1) This problem is an easy place to make sign errors.

*Applications, Trigonometric Tools: Powers and Roots of Complex Numbers*

**In-Text Examples**

1) Students may actually be able to skip the step of calculating the three roots if they realize that the roots are evenly spaced about a circle and are able to figure out how to graph them based on that knowledge. This probably shouldn’t be penalized, as it demonstrates understanding of the principles behind \begin{align*}n^{\mathrm{th}}\end{align*} roots.

2) The bit about using the Pythagorean Theorem and polar coordinates is somewhat of a red herring; those can be used to find other values in the given diagram, but to find L, students need only use the Law of Cosines.

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