10.5 Chapter Summary and Review

Key Concepts

Note
Polar Coordinates
The polar coordinates of a point $P$ in the plane are $(r, θ)$ , where
- $| r |$ is the distance from to the pole,
- $θ$ is the angle measured counterclockwise from the polar axis to the ray through $P$ from the pole.
Note
Non-Uniqueness of Polar Coordinates
1. Any point with polar coordinates $(r, θ)$ also has coordinates $open parenthesis r, θ plus 2 k π close parenthesis$ , where $k$ is an integer.
2. The point $(r, θ)$ can also be designated by $(- r, θ + π)$ .
3. The pole has coordinates $open parenthesis 0, θ close parenthesis$ , for any value of $θ$ .
In the polar plane, the coordinate grid lines are circles centered at the pole, with equations $r = k$ , and lines through the pole, with equations $θ = k$ .
Note
Conversion Equations
1. To convert from polar coordinates $(r, θ)$ to Cartesian:
  $\begin{aligned} x & = r \cos (θ) \\ y & = r \sin (θ) \end{aligned}$
2. To convert from Cartesian coordinates $(x, y)$ to polar:
  $\begin{aligned} r = \sqrt{x^{2} + y^{2}} \\ \tan (θ) = \frac{y}{x} \end{aligned}$
where the choice of $θ$ depends on the quadrant.
To convert an equation from Cartesian to polar coordinates, we replace each $x$ with $r \cos (θ)$ and each with $y$ with $r \sin)$ . To convert an equation from polar to Cartesian coordinates, look for expressions of the form $r \cos (θ), r \sin (θ), r^{2}$ , or $\tan (θ)$ .
When graphing an equation in polar coordinates, we think of sweeping around the pole in the counterclockwise direction, and at each angle $θ$ the $r$ -value tells us how far the graph is from the pole.
Standard graphs in polar coordinates include circles and roses, cardioids and limaçons, lemniscates, and spirals.
To find the intersection points of the polar graphs $r = f (θ)$ and $r = g (θ)$ we solve the equation $f (θ) = g (θ)$ . In addition, we should always check whether the pole is a point on both graphs.
Note
Imaginary Unit
We define the imaginary unit, $i$ , by
$i^{2} = - 1 or i = \sqrt{- 1}$
The square root of a negative number is an imaginary number: if $a > 0, \sqrt{- a} = i \sqrt{a}$
A complex number $z$ is the sum of a real number and an imaginary number, $z = a + b i$ .
We can perform the four arithmetic operations on complex numbers.
Note
Operations on Complex Numbers
$z_{1} + z_{2} = (a + b i) + (c + d i) = (a + c) + (b + d) i$
$z_{1} - z_{2} = (a + b i) - (c + d i) = (a - c) + (b - d) i$
$z_{1} z_{2} = (a + b i) (c + d i) = (a c - b d) + (a d + b c) i$
$\frac{z_{1}}{z_{2}} = \frac{a + b i}{c + d i} = \frac{a + b i}{c + d i} \cdot \frac{c - d i}{c - d i} = \frac{a c + b d}{c^{2} + d^{2}} + \frac{b c - a d}{c^{2} + d^{2}} i$
The product of a nonzero complex number and its conjugate is always a positive real number.
$z \bar{z} = (a + b i) (a - b i) = a^{2} - b^{2} i^{2} = a^{2} - b^{2} (- 1) = a^{2} + b^{2}$
.
We can graph complex numbers in the complex plane.
We can visualize the sum of two complex numbers by vector addition in the complex plane.
Note
Fundamental Theorem of Algebra
Let $p (x)$ be a polynomial of degree $n greater than or equal to 1$ . Then $p (x)$ has exactly $n$ complex zeros.
The nonreal zeros of a polynomial with real coefficients always occur in conjugate pairs.
Multiplying a complex number by $i$ rotates its graph by $90 degrees$ around the origin.
Note
Polar Form for a Complex Number
The complex number $z = a + b i$ can be written in the polar form
$z = r (\cos (θ) + i \sin (θ))$
where
$r = \sqrt{a^{2} + b^{2}}$
and $θ$ is defined by
$a = r \cos (θ), b = r \sin (θ), 0 \leq θ \leq 2 π$
The angle $θ$ is called the argument of the complex number, and $r$ is its length, or modulus.
Note
Product and Quotient in Polar Form
If $z_{1} = r (\cos α + i \sin α)$ and $z_{2} = R (\cos β + i \sin β)$ , then
$z_{1} z_{2} = r R (\cos (α + β) + i \sin (α + β))$
and
$\frac{z_{1}}{z_{2}} = \frac{r}{R} (\cos (α - β) + i \sin (α - β))$
Note
De Moivre's Theorem
If $z = r (\cos (α) + i \sin (α))$ is a complex number in polar form, and $n$ is a positive integer, then
$z^{n} = r^{n} (\cos (n α) + i \sin (n α))$
Note
Roots of a Complex Number
A complex number $z = r (\cos (α) + i \sin (α))$ in polar form has $n$ complex $n$ th roots, given by
$z_{k} = r^{1 / n} (\cos (\frac{α + 2 π k}{n}) + i \sin (\frac{α + 2 π k}{n}))$
for $k = 0, 1, 2, \dots, n - 1$ .

Review Problems

For Problems 1–4, use the grid at right to plot the points whose polar coordinates are given.

$(4, \frac{5 π}{6})$

$(1, \frac{- 5 π}{3})$

$(- 3, \frac{π}{2})$

$(2, \frac{7 π}{4})$

For Problems 5–8, convert the polar coordinates to Cartesian coordinates.

$(1, \frac{5 π}{4})$

$(\frac{- \sqrt{2}}{2}, \frac{- \sqrt{2}}{2})$

$(0, \frac{π}{12})$

$open parenthesis 3.4, minus 1.5 close parenthesis$

$open parenthesis 0.241, minus 3.391 close parenthesis$

$(- 5.6, - 1.1)$

For Problems 9–12, convert the Cartesian coordinates to polar coordinates with $r greater than or equal to 0$ and $0 less than or equal to θ less than or equal to 2 π$ . Give exact values for $r$ and $θ$ .

$open parenthesis minus 3, 3 close parenthesis$

$(3 \sqrt{2}, \frac{3 π}{4})$

$open parenthesis 0, minus 2 close parenthesis$

$open parenthesis 5, minus 2 close parenthesis$

$(\sqrt{29}, \tan^{- 1} (\frac{- 2}{5}) + 2 π)$

$(- 15, - 8)$

For Problems 13–16, sketch the region described by the inequalities.

$r \geq 0, \frac{- π}{4} \leq θ \leq \frac{π}{4}$

$1 less than or equal to r less than or equal to 3, 0 less than or equal to θ less than or equal to π$

$0 less than or equal to r less than or equal to 2$

$r greater than or equal to 4$

For Problems 17–20, convert the equation into Cartesian coordinates.

$r equals 1$

$x^{2} + y^{2} = 1$

$r = - 3 \sec (θ)$

$r = \frac{6}{1 - 2 \cos (θ)}$

$x^{2} + y^{2} = (2 x + 6)^{2}$

$3 \tan (θ) = 6 r \sin (θ) - 1$

For Problems 21–24, convert the equation into polar coordinates.

$x + y = 2$

$r \cos (θ) + r \sin (θ) = 2$

$\sqrt{x^{2} + y^{2}} = 4 y$

$\frac{y}{x} = \sqrt{x^{2} + y^{2}}$

$\tan (θ) = r$

$y^{2} = 2 y = x - x^{2}$

For Problems 25–28, use the catalog of polar graphs to help you identify and sketch the curve. Check your work by graphing with a calculator.

$r equals 3$

Circle of radius 3 centered at the origin

$θ = \frac{3 π}{4}$

$r equals 6 cos ⁡ open parenthesis θ close parenthesis$

Circle of radius 3 centered at $open parenthesis 3, 0 close parenthesis$

$r squared equals 9 sin ⁡ open parenthesis 2 θ close parenthesis$

For Problems 29–32, write a polar equation for the graph.

$r equals 4$

$r equals 4 cos ⁡ open parenthesis θ close parenthesis$

For Problems 33–36, find the coordinates of the intersection points of the two curves analytically. Then graph the curves to verify your answers.

$r = 3 + 2 \sin (θ), r = 4$

$(4, \frac{π}{6})$ , $(4, \frac{5 π}{6})$

$r = 3 \cos (θ), r = \sqrt{3} \sin (θ)$

$r = 4 \sin (θ), r = - 4 \cos (θ)$

$(2 \sqrt{2}, \frac{3 π}{4})$ and the pole

$r = 2 + 6 \sin (θ), r = 4 \sin (θ)$

For Problems 37–40, perform the indicated operations on the complex numbers.

$\frac{5 - 10 i}{2 - i}$

$4 minus 3 i$

$(4 - 7 i) (1 + i)$

$5 i (2 - i) - (7 + 6 i)$

$- 2 + 4 i$

$- 8 + 3 i + \frac{9 - 4 i}{i}$

For Problems 41–44, evaluate the polynomial for the given values of the variable.

$z^{2} + 4 z + 6$

$z = - 2 + i$
$z = - 2 - 1$

$z^{2} - 6 z + 12$

$z = 3 - 2 i$
$z = 3 + 21$

$3 w^{2} - 18 w + 31$

$w = 3 + 4 i$
$w = 3 - 4 i$

$minus 44$
$minus 44$

$2 w^{2} + 8 w + 11$

$w = - 2 - 5 i$
$w = - 2 + 51$

Verify that $z_{1} = 2 + i$ and $z_{2} = 2 - i$ are roots of the equation $x^{2} - 4 x + 5 = 0$ .

$(2 \pm i)^{2} - 4 (2 \pm i) + 5 = (4 \pm 4 i - 1) - (8 \pm 4 i) + 5 = 0$

Verify that $z_{1} = - 3 + 4 i$ and $z_{2} = - 3 - 4 i$ are roots of the equation $x^{2} + 6 x + 25 = 0$ .

For Problems 47–50, expand the product of polynomials.

$[z - (- 2 + i)] [z - (- 2 - i)]$

$z^{2} + 4 z + 5$

$[w - (1 + 3 i)] [w - (1 - 3 i)])]$

$[s + (5 + 4 i)] [s + (5 - 4 i)]$

$s^{2} - 10 s + 41$

$[x + (- 6 + i)] [x + (- 6 - i)]$

For Problems 51–52, sketch the set of points in the complex plane.

$z_{1} = - 3 + 2 i,$ $z_{2} = - 3 - 2 i,$ $z_{3} = 3 + 2 i,$ $z_{4} = 3 - 2 i$

$w_{1} = 4 - 6 i,$ $w_{2} = 4 + 6 i,$ $w_{3} = - 4 - 6 i,$ $w_{4} = - 4 + 6 i$

For Problems 53–56,

Given one solution of a quadratic equation with rational coefficients, find the other solution.
Write a quadratic equation that has those solutions.

$- 1 + 7 i$

$- 1 - 7 i$
$x^{2} + 2 x + 50 = 0$

$2 minus 5 i$

$3 - \sqrt{2} i$

$3 + \sqrt{2} i$
$x^{2} - 6 x + 11 = 0$

$4 + \sqrt{3} i$

For Problems 57–60, write the complex numbers in standard form. Give exact values for your answers.

$10 (\cos (\frac{- π}{6}) + i \sin (\frac{- π}{6}))$

$5 \sqrt{3} - 5 i$

$8 (\cos (\frac{5 π}{4}) + i \sin (\frac{5 π}{4}))$

$5 \sqrt{2} (\cos (\frac{π}{4}) + i \sin (\frac{π}{4}))$

$5 plus 5 i$

$6 \sqrt{3} (\cos (\frac{- π}{3}) + i \sin (\frac{- π}{3}))$

For Problems 61–66, write the complex numbers in polar form. Give exact values for your answers.

$3 minus 3 i$

$3 \sqrt{2} (\cos (\frac{7 π}{4}) + i \sin (\frac{7 π}{4}))$

$- 4 - 4 i$

$minus 5$

$5 open parenthesis cos ⁡ open parenthesis π close parenthesis plus i sin ⁡ open parenthesis π close parenthesis close parenthesis$

$minus 7 i$

$- 1 - \sqrt{3} i$

$2 (\cos (\frac{4 π}{3}) + i \sin (\frac{4 π}{3}))$

$6 + 2 \sqrt{3} i$

For Problems 67–70, find the product $z sub 1 z sub 2$ and the quotient $\frac{z_{1}}{z_{2}}$ .

$z_{1} = 8 (\cos (\frac{π}{6}) + i \sin (\frac{π}{6}))$ $z_{2} = 2 (\cos (\frac{5 π}{6}) + i \sin (\frac{5 π}{6}))$

$\begin{aligned} z_{1} z_{2} & = 16 (\cos (π) + i \sin (π)) \\ = - 16, \frac{z_{1}}{z_{2}} \\ = 4 (\cos (\frac{- 2 π}{3}) + i \sin (\frac{- 2 π}{3})) \\ = - 2 - 2 \sqrt{3} i \end{aligned}$

$z_{1} = 9 (\cos (\frac{- 2 π}{3}) + i \sin (\frac{- 2 π}{3}))$ $z_{2} = 3 (\cos (\frac{π}{3}) + i \sin (\frac{π}{3}))$

$z_{1} = 5 (\cos (\frac{- 7 π}{12}) + i \sin (\frac{- 7 π}{12}))$ $z_{2} = \frac{1}{2} (\cos ((\frac{π}{4}) + i \sin (\frac{π}{4})))$

$\begin{aligned} z_{1} z_{2} & = \frac{5}{2} (\cos (\frac{- π}{3}) + i \sin (\frac{- π}{3})) \\ = \frac{5}{4} - \frac{5 \sqrt{3}}{2} i \end{aligned},$ $\begin{aligned} \frac{z_{1}}{z_{2}} & = 10 (\cos (\frac{- 5 π}{6}) + i \sin (\frac{- 5 π}{6})) \\ = - 5 \sqrt{3} - 5 i \end{aligned}$

$z_{1} = 14 (\cos (\frac{3 π}{2}) + i \sin (\frac{3 π}{2}))$ $z_{2} = 2 (\cos \frac{π}{6} + i \sin \frac{π}{6})$

For Problems 71–74, find the power.

${(\frac{\sqrt{3}}{2} + \frac{1}{2} i)}^{12}$

$1$

${(1 - \sqrt{3} i)}^{6}$

${(- \sqrt{5} + \sqrt{5} i)}^{- 4}$

$\frac{- 1}{100}$

$(- 1 - i)^{- 8}$

For Problems 75–78,

Find the roots and plot them in the complex plane.
Write the roots in standard form.

The square roots of $minus 16 i$

$- 2 \sqrt{2} + 2 \sqrt{2} i$ , $2 \sqrt{2} - 2 \sqrt{2} i$

The cube roots of $minus 8$

The cube roots of $minus 27 i$

$3 i$ , $\frac{- 3 \sqrt{3}}{2} - \frac{3}{2} i$ , $\frac{3 \sqrt{3}}{2} - \frac{3}{2} i$

The square roots of $- 2 + 2 \sqrt{3} i$

For Problems 79–82, solve the equation.

$z^{6} + 27 = 0$

$3 open parenthesis cos ⁡ θ plus i sin ⁡ θ close parenthesis,$ for $θ = \frac{π}{6},$ $\frac{π}{2},$ $\frac{5 π}{6},$ $\frac{7 π}{6},$ $\frac{3 π}{2},$ $\frac{11 π}{6}$

$z^{4} - 6 z^{2} + 12 = 0$

$z^{4} + 2 z^{2} + 4 = 0$

$\sqrt{2} (\cos (θ) + i \sin (θ)),$ for $θ = \frac{π}{3},$ $\frac{2 π}{3},$ $\frac{4 π}{3},$ $\frac{5 π}{3}$

$z^{6} - 8 = 0$

Trigonometry by Katherine Yoshiwara (yoshiwarabooks.org), GNU Free Documentation License 1.2 or later. Adapted for the XYZ HTML edition with the authors' permission (recorded 2026-07-04). License: GFDL-1.2-or-later.