# 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 *not* another way. In general, if we change the sign of the

(Here’s part of the reason why: When we change the sign of the

**Review Questions**

2) Students’ answers should include the original coordinates

*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

**Review Questions**

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

2) The rose will have

*Applications, Trigonometric Tools*

**In-Text Examples**

2) Students may think that the total distance

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

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

**Review Questions**

1) Students working in degrees instead of radians will get

*Conic Section Transformations*

**In-Text Examples**

1) Mixing up the

5) You may want to point out that plugging in a negative value for

**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

*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

## Systems of Polar Equations

*Graph and Calculate Intersections of Polar Curves*

**In-Text Examples**

3) Students may be very confused by the idea that

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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