Chapter 6: The Polar System
Chapter Outline
 6.1. Polar Coordinates
 6.2. Graphing Basic Polar Equations
 6.3. Converting Between Systems
 6.4. More with Polar Curves
 6.5. The Trigonometric Form of Complex Numbers
 6.6. The Product & Quotient Theorems
 6.7. De Moivre’s and the nth Root Theorems
Chapter Summary
Chapter Summary
In this chapter we made the connection between complex numbers and trigonometry. First, we started with the polar system, by graphing and converting equations into polar coordinates. This allowed us to compare the complex plane with the polar plane and we realized that there are many similarities. Because of this, we are able to convert complex numbers into polar, or trigonometric, form. Converting complex numbers to polar form makes it easier to multiply and divide complex numbers by using the Product and Quotient theorems. These theorems lead to De Moivre’s Theorem, which is a shortcut for raising complex numbers to different powers. Finally, we were able manipulate De Moivre’s Theorem to find all the complex solutions to different equations.
Vocabulary
 Argument

In the complex number
r(cosθ+isinθ) , the argument is the angleθ .
 Modulus

In the complex number
r(cosθ+isinθ) , the modulus isr . It is the distance from the origin to the point(x,y) in the complex plane.
 Polar coordinate system
 A method of recording the position of an object by using the distance from a fixed point and an angle consisting of a fixed ray from that point. Also called a polar plane.
 Pole
 In a polar coordinate system, it is the fixed point or origin.
 Polar axis
 In a polar coordinate system, it is the horizontal ray that begins at the pole and extends in a positive direction.
 Polar coordinates

The coordinates of a point plotted on a polar plane
(r,θ) .
 Polar Equation
 An equation which uses polar coordinates.
 Polar Form

Also called trigonometric form is the complex number
x+yi written asr(cosθ+isinθ) wherer=x2+y2−−−−−−√ andtanθ=yx .
Review Questions
 Plot
A(−3,3π4) and find three other equivalent coordinates.  Find the distance between
(2,94∘) and(7,−73∘) .  Graph the following polar curves.

r=3sin5θ 
r=6−3cosθ 
r=2+5cos9θ

 Determine the equations of the curves below.
 Convert each equation or point into polar form.

A(−6,11) 
B(15,−8) 
C(9,40) 
x2+(y−6)2=36

 Convert each equation or point into rectangular form.

D(4,−π3) 
E(−2,135∘) 
r=7 
r=8sinθ

 Determine where
r=6+5sinθ andr=3−4cosθ intersect.  Change
−3+8i into polar form.  Change
15∠240∘ into rectangular form.  Multiply or divide the following complex numbers using polar form.

(7cis7π4)⋅(3cisπ3) 
8∠80∘2∠−155∘

 Expand
[4(cosπ4+isinπ4)]6  Find the
6th roots of 64 and graph them in the complex plane.  Find all the solutions of
x4+32=0 .
Review Answers

A(−3,3π4) three equivalent coordinates→(3,−π4),(3,7π4),(−3,−5π4) . 
(2,94∘) and(7,−73∘) d=22+72−2(2)(7)cos(94∘−(−73∘))−−−−−−−−−−−−−−−−−−−−−−−−−−−√=4+49−28cos167∘−−−−−−−−−−−−−−−√=80.28−−−−√≈8.96  a)
r=3sin5θ b)r=6−3cosθ c)r=2+5cos9θ 
r=2−6cosθ 
r=7+3sinθ

 a.
A(−6,11)→r=36+121−−−−−−−√≈12.59,tanθ=−116,θ=118.6∘→(12.59,118.6∘) b.B(15,−8)→r=225+64−−−−−−−√=17,tanθ=−815,θ=−28.1∘→(17,−28.1∘) c.C(9,40)→r=91+1600−−−−−−−−√=41,tanθ=409,θ=77.3∘→(41,77.3∘) d.x2+(y−6)2r2cos2θ+(rsinθ−6)2r2cos2θ+r2sin2θ−12rsinθ+36r2−12rsinθr2r=36=36=36=0 or=12rsinθ=12sinθ  a.
D(4,−π3)→x=4cos(−π3)=2,y=4sin(−π3)=−23√→(2,−23√) b.E(−2,135∘)→x=−2cos135∘=2√,y=−2sin135∘=−2√→(2√,−2√) c.r=7→r2=49→x2+y2=49 d.rr2x2+y2y2−8yy2−8y+16(y−4)2x2+(y−4)2=8sinθ=8rsinθ=8y=−x2=16−x2=16−x2=16 
r=6+5sinθ andr=3−4cosθ ∗ angle measures in the graph are in radians Note: The two determined points of intersection [(6.91,2.95) and(1.44,−1.16) ] were estimated from the trace function on a graphing calculator and are not precise solutions for either equation. 
−3+8i,x=−3,y=8→r=(−3)2+82−−−−−−−−−√≈8.54,tanθ=−83→θ=110.56∘ 8.54(cos110.56∘+isin110.56∘ 
15∠240∘,r=15,θ=240∘→x=15cos240∘=−7.5,y=15sin240∘=−153√2=−7.53√ So,15∠240∘=−7.5−7.5i3√ .
(7cis7π4)⋅(3cisπ3)=21cis(7π4+π3)=21cis25π12 
8∠80∘2∠−155∘=4∠(80∘−(−155∘))=4∠235∘


[4(cosπ4+isinπ4)]6=46(cos6π4+isin6π4)=4096(cos3π2+isin3π2)  64 in polar form is
64(cosπ−isinπ) Graph of the solutions:[64(cos(π+2πk)+isin(π+2πk))]162(cos(π+2πk6)+isin(π+2πk6))2(cos(π6+πk3)+isin(π6+πk3))z1=2(cos(π6+0π3)+isin(π6+0π3))=2cosπ6+2isinπ6=23√2+2i2=3√+iz2=2(cos(π6+π3)+isin(π6+π3))=2cosπ2+2isinπ2=2iz3=2(cos(π6+2π3)+isin(π6+2π3))=2cos5π6+2isin5π6=−23√2+2i2=−3√+iz4=2(cos(π6+π)+isin(π6+π))=2cos7π6+2isin7π6=−23√2−2i2=−3√−iz5=2(cos(π6+4π3)+isin(π6+4π3))=2cos3π2+2isin3π2=−2iz6=2(cos(π6+5π3)+isin(π6+5π3))=2cos11π6+2isin11π6=23√2−2i2=3√−i 
x4+32=0→x4=−32+0i=−32(cosπ+isinπ)[32(cos(π+2πk)+isin(π+2πk))]1422√4(cos(π+2πk4)+isin(π+2πk4))22√4(cos(π4+πk2)+isin(π4+πk2))z1=22√4(cos(π4)+isin(π4))=22√4cosπ4+2i2√4sinπ4=22√42√2+2i2√42√2 =2√43+i2√43z2=22√4(cos(π4+π2)+isin(π4+π2))=22√4cos3π4+2i2√4sin3π4=−22√42√2+2i2√42√2 =−2√43+i2√43z3=22√4(cos(π4+π)+isin(π4+π))=22√4cos5π4+2i2√4sin5π4=−22√42√2−2i2√42√2=−2√43−i2√43z4=22√4(cos(π4+3π2)+isin(π4+3π2))=22√4cos7π4+2i2√4sin7π4=22√42√2−2i2√42√2=2√43−i2√43
Texas Instruments Resources
In the CK12 Texas Instruments Trigonometry FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9704.