美国数学美国数学311 Polar and Complex

    11-1: Polar Coordinates and Complex Numbers   Important things to remember!!  Converting from polar to rectangular: 1) x = r cos 0 2) y = r sin 0 Converting from rectangular to polar:  3) r2 = x2 + y2 4) tan 0 = (y/x) Click for polar graphing calculator! (Manipula Math)   Example problems 1) Change (3, 4) to polar coordinates. Solution: Using property 3 from above, find r.   32 + 42 = 25 and take the square root.                 Therefore, r = 5 Using property 4 from above, tan 0 = (4/3). Use your calculator set to degree mode, the answer is: 53.1 degrees. (rounded to nearest tenth.                 Therefore, the point is ( 5, 53.1o) 2) Change ( 4, 150o) to rectangular coordinates Solution: Using property 1 from above x = 4 cos 150. Using your calculator you get x = -3.46 rounded to hundredths. Using property 2 from above y = 4 sin 150. Using your calculator you get y = 2              Therefore, the point is ( -3.46, 2)  3) Change (-3, -7) to polar coordinates Solution: Using property 3 from above find r. Square -3 and add to the square of -7 you get 58 and taking the square root on your calculator means r = 7.6   Piece of cake! To find theta we use property 4, tan 0 = 7/3. Notice I used positive values! I did this to find the reference angle in quad I. The reference angle is equal to 66.8 degrees. Since we are in quadrant III, (look at the signs of the original problem) we should add 180. Why? ( we are in quad III and we memorized we add 180 in that quad, right?) So the angle is 246.8o.             Therefore, the point is (7.6, 246.8o) ( all to simple!!!)    Note You can use the special functions on the TI-82 to find all the above answers.  Make sure your calculator is set in degree mode(if you want the answer in degrees) and press 2nd function angle.  This takes you to a menu that has choice 5 and 6 to change to polar form.  Choice 5 gives you the radius and choice 6 gives you the theta value.  Example (1,3) Get to choice 5 on your calculator: You should see this R arrow Pr ( you type 1, 3) and the calculator gives you 3.16227766. This is r.  Round it and your all set.  Then go back to the angle menu and choose 6.  You should see R arrow P0 type 1,3) again and get 71.56505118 round this and the point would be ( 3.2, 71.6o)    (Yikes stripes this easy!!) Do basically the same thing to find x and y by using choices 7 and 8. If the problem is in degrees set calc to degrees and if in rads make sure calc is in rads.    Sketching graphs in polar form using the TI-82 Graphs are sketched at the bottom of the page! To use your graphing calculator is extremely easy to graph in polar form.  First, get to the mode screen.  Make sure you highlight the radian button and drop down one line and highlight the pol button.  Now hit the y= key. Instead of y= it's now r= !!  Amazing what modern technology will do!!  When you hit the x, t, theta button, it now shows a theta!!  Double wowie! Try a few problems 1) r = 3 sin 0 Go ahead, type it in and hit graph.  Did you get something that sort of looked like a circle?  Hit zoom then choice 5.  Look better?  It should be a circle tangent at the origin and translated up.  The top point is (0, 3) 2) r = 4 cos 0 Try this one.  Hit y= and type it in.  Hit graph.  This time, you should get a circle tangent again to the origin but moved to the right.  The point farthest right is (2, 0).  Wonder if this is a pattern?  Sine up or down and cosine right or left? 3) r = - 5 sin 0    Circle moved down 4) r = -3 cos 0    Circle moved left. 5) r = 3 sin 40  Interesting graph.   Looks like 8 leaves on a rose.  Hit zoom and take choice 1. Move the cursor to a corner of the picture and hit enter.  Hit the down arrow and make the line bigger than the picture.  Hit enter when it is.  Now stretch it across until it makes a box around the picture.  Hit enter  Now you see a better picture of it.  The number of leaves depend on the number multiplying 0.  If it is odd that's how many leaves you get.  If it is even like this one, you get double the number of leaves.  Let's see, the number for this problem was 4, I double it and I should have 8 leaves.   Count them on the screen.   How about that.  To be a picture like this, the number multiplying 0 must be bigger than 1 and a whole number.  6) r = 4 cos 30    Same as above with 3 leaves.  7) r = 3 cos 20   Another leaved rose.  This time 4 leaves.  Remember after each problem to reset the calculator to standard zoom by hitting zoom then 6  8) r = 2 + 2 cos 0.  A cardiod.  (heart- shaped)  9) r = 1 + 2 sin 0    A limicon  10) r = sec 0      What is this?   A vertical line if you remember sec is 1/cos   Graphs  ..........     ..........   ..........    ..........   .......... Let's head on toward the next section.  Geometric representation of complex numbers!! Sounds like fun!         11-2 Geometric and Trigonometric Representation of Complex Numbers                                                               Imaginary axis real axis The above diagram is an Argand diagram.  Notice that the real numbers are on the x-axis and the imaginary numbers are on the y-axis.  Finding imaginary numbers in this plane are as easy as finding points in the real plane.  In the form a + bi, a is the real part and b is the imaginary part.  Move a units right or left ( depending on + or  - ) and b units up or down (depending on + or -). Rectangular form: z = a + bi Polar form: z = r cos 0 + (r sin 0)i (remember a = r cos 0 and b = r sin 0 substitute these in and presto!) factor out the r and get: z = r(cos 0 + i sin 0) Math shorthand looks like this: z = r cis 0 The absolute value of z (the distance from any point to the origin) =               _______ | z | = \/ a2 + b2   (good old Pythagorus again) Example problems: 1) Express 3 cis 50o in rectangular form.     Solution: using the fact that a = r cos 0 and b = r sin 0 implies that                    a = 3 cos 50 and b= 3 sin 50. Using your calculator gets us                    a = 1.93 and b = 2.30 with both answers rounded to hundredths.                 Therefore the answer is: 1.93 + 2.30i 2) Express  -1 -2i in polar form.                                                           __________         __     Solution: Use the fact that r = \/(-1)2 + (-2)2 r = \/ 5  = 2.24.  Now use the fact that the Tan 0 = (y/x) (see page 1 if you forgot!!).   0 = tan-1(2/1).  Using your calculator gives us 63.4o.  Add 180 (why? we are in the 3rd quadrant!!!)  63.4 + 180 = 243.4                   Therefore, the answer is: 2.24 cis 243.4o To Multiply two complex numbers in polar form: 1) Multiply their absolute values 2) Add their polar angles. In math terms if z1 = r cis w and z2 = s cis y then z1z2 = rs cis ( w + y) Example problem: 1) Express (5cis 30o)(7cis60o) in polar and rectangular form.     Solution:  Polar form first: multiply the radii and add the angles.                     Answer in polar form: 35 cis 90o         Now change this to get the rectangular form. Remember, a = r cos 0 and b = r sin 0 so a = 35 cos 90 = 0 and b = 35 sin 90 = 35. Answer in rectangular form is: 0 + 35i Next section involves finding the powers of complex numbers!!           11 - 3 Powers of Complex Numbers   Demo: Snail Shell (Manipula Math) Important fact to remember De Moivre's Theorem says If z = r cis 0 then zn = r n cis n0 This formula allows us to make some very cool graphs using argand diagrams. A lot of these graphs emulate many of the spirals and figures that we have in nature. Check out the book on page 409 as an example. Pretty cool. Here's how to calculate them.   Example problem: Find the powers for n = 1, 2, 3, 4, 5, 6 and make an argand diagram for each. z = 1 + i Solution: Find each power first.                                         _______ z = 1.4 cis 45 ( where did I get this? Remember, r = \/a2 + b2 and we can find the angle from tan 0 = (y/x) r2 = (1.4)2 cis [(2)(45)] = 2 cis 90     (1.4 is the square root of two silly) z3 = (1.4)3 cis [(3)(45)] = 2.8 cis 135 z4 = (1.4)4 cis [(4)(45)] = 4 cis 180 z5 = (1.4)5 cis [(5)(45)] = 5.6 cis 225 z6 = (1.4)6 cis [(6)(45)] = 8 cis 270     If you make an argand diagram by sweeping out each angle and going out the appropriate distance, you should end up with a spiral similar to a snail shell. Notice that each angle increases by 45 degrees and each radius gets increasingly larger thus forming a spiral. If you play connect the dots with the tips of each answer, it forms a simple snail shell. The pattern is endless as the higher powers keep increasing in size.  Look at the argand diagram for the above:      That's about all for this section. Not to bad, but the numbers can make the arithmetic a little sticky.   Have your calculator handy!!        To continue to the last section 11.4 hit       To back up and regroup in section 11.2 hit      11 - 4 Roots of Complex Numbers Try the quiz at the bottom of the page! go to quiz   We already know from Algebra II that every equation of power 2 has 2 solutions and every equation of power 3 has 3 solutions etc, etc ad nauseum.  But up until now, we have only one solution for the equation x3 = 8.  The only one you could find was x =2.  Where are the other two?  We know they exist, but how do you find them.  If we extend De Moivre's to find roots, suprise!!  We can find the missing roots!!  How you might ask?  Good question!  I guess I will show you!  Hang on to your hat and off we go! Important definition ( This means pay attention!) The n nth roots of z = r cis 0 are: z1/n = r1/n cis ( 0/n + k . 360o /n) for k = 0, 1, 2, 3, ... n-1 Say what?  What are you trying to tell me?  Here's the deal!  Say you want to find the 3 cubed roots of 8.  That is x3 = 8.  Change 8 into r cis 0.  You remember that right! 8 is on the x-axis so r cis 0 = 8 cis 0o.  Since we are working with the third power, k will equal 0, 1, and 2.  Look at the formula!  The highest k goes to is n-1.  Tada!  Now plug and chug!  One at a time. When k = 0, z1/3 = 81/3 cis ( 0/3 + 0 . 360/3) = 2 cis( 0 + 0) = 2cis 0 = 2cos 0 + 2i sin 0 = 2      Voila! When k=1, z1/3 = 81/3 cis ( 0/3 + 1. 360/3) = 2 cis ( 0+120) = 2cis 120 =                                                        __ 2cos120 + 2i sin 120 = -1 + i \/ 3  .   How about that!  An imaginary root.  Imagine that( pun intended) When k=2, z1/3 = 81/3 cis ( 0/3 + 2 . 360/3) = 2 cis(0 + 240) = 2cis 240 =                                                  __ 2cos240 + 2i sin 240 = -1 - i\/ 3  .  Another imaginary number.  Hey, I seem to recall something about these guys showing up in pairs?   Do you? So how do you go about proving what you really have are the roots of 8. How do you check? Any brilliant ideas? That's right, you multiply. What?  You heard me, Multiply. Proof: 2 x 2 x 2 = 8 ( boy that was tough!)                                         __            __            __  Proof: (-1 -i\/ 3)(-1 -i \/ 3)(-1 -i \/3 ) You guessed it FOIL!                                     __              __ (-2 + 2i \/ 3)( -1 - i \/ 3) = 8 (Ripleys believe it or not!)                                                        __              __              __ Proof: (-1 + i \/ 3)(-1 + i \/ 3)( -1 + i \/ 3) =                                       __              __ (-2 - 2i \/ 3)( -1 - i \/ 3) = 8 ( believe it if you must) Basically, if you can do a couple of these monsters, you are doing pretty well. Thus ends chapter 11 and our glorious trek through the cyberspace of trigonometry. But stay tuned, it's not quite over. Up next is what you have waited for with baited breath! The sample test. To go onto the sample test hit : Sample test To back up and review before disaster strikes hit: Previous page 窗体顶部 窗体底部 HTMLCONTROL Forms.HTML:Hidden.1 HTMLCONTROL Forms.HTML:Hidden.1 HTMLCONTROL Forms.HTML:Hidden.1 HTMLCONTROL Forms.HTML:Hidden.1   Current quizaroo #  11   1)  Find the polar form of the point (3, 4).  Round  the angle to tenths place.   a)  (4, 45o) b)  (3, 60o) c)  (5, 45o) d)  (5, 53.1o) e)  (4, 53.1o)       2)  r = 4 sin 4  is an example of what type of graph?            a)  4-leaved rose    b)  limacon c)  8-leaved rose d)  cardioid e)  lemniscate       3)  Write (-3 + 4i) in polar form. a)  5cis 53.1o b)  5cis 126.9o c)  4cis 53.1o d)  3cis 126.9o e)  5cis 233.1o       4)  Give the polar form of (2cis 45o)3 a)  8cis 135o b)  8cis 45o c)  2cis 45o d)  2cis 135o e)  6cis 45o       5)  Find the three cubed roots of 27.          a)  3, 3i, -3i                            __ b)  3, (-3 + 3i \/ 3 )/2                          __ c)  3, (-1 + i \/ 3 )                     __ d)  3, -3, \/ 3                            __ e)  3, (-3 + 3i \/ 3 )        click here for answers!! Chapter 11 practice test. Name___________________________ 1) Plot each point in polar coordinates. Then name two other points one a positive value for r and one a negative value for r. a) (5, 75o) b) (4, 3/4) 2) Change the following rectangular coordinates to polar coordinates where 0 is in degrees. a) ( -3, 4) b) (-4, 0) 3) Change the following rectangular point to polar form where 0 is in radians. a) ( 5, 13) 4) Change the following polar points to rectangular coordinates. a) (7, 240o) b) ( 3, 142o) 5) Using your graphing calculator reproduce the following graphs. a) r = -5sin 0 b) r = 4 cos 0 c) r = 3 - 2 cos 0 d) r = 2 + 3 sin 0 e) r = 3cos 30 f) r = 5 sin 40 g) r = 20 6) Express 3 + 2i in polar form. 7) Express 5 cis 320o in rectangular form. 8) If z1 = 3 + 5i and z2 = 5 + 6i find the product: a) in cis form b) change to rectangular form c) draw an argand diagram for z1, z2, and z1z2. 9) Give the polar form for ( 3 cis 60o)2 10) let z = 1 + i. a) Expess z and z6 in polar form. b) Express z6 in rectangular form. 11) Find the 3 cubed roots of 1. That's right. One has 3 cubed roots. Amazing. Answers and how they are worked out will be posted in a couple of days. This test is a little longer than the actual one, but is representative of an actual test. Questions have been changed to protect the innocent. I seem to remember that phrase from some old tv show. Trivia question. Which show? Hope this test will be helpful in preparing you for your opportunity.  Next postings will be for chapter 13. Arithmetic and Geometric Sequences. This page hosted by  Get your own Free Home Page Let me see the answers, I have worked on these guys forever: hit ---> Answers this way Man, I need some more help before I check them, send me back to study ----> Need more help   1a) (5, 435o) I added 360o 1a) (-5, 255o) I added 180o 1b) (4, 11/4) I added 2 1b) (-4, 7/4) I added   a) r = \/ 9 + 16 = 5and tan This is the reference angle. The angle is supposed to be in the 2nd quadrant. Subtract from 180. Answer is   (5, 126.9o) 2b) Easy one. r = 4 and  = 180o  Why? Answer is: (4, 180o)  r = \/ 25 + 169 = 13.9 and Tan(13/5) = 1.21 in rads. Answer is: (13.9, 1.21)  4a) x = 7cos 240o = -3.5 and y = 7sin 240o = -6.1 Answer is: (-3.5, -6.1)  4b) x = 3cos 142o = -2.4 and y = 3sin 142o = 1.8 Answer is: (-2.4, 1.8)  5a)  5b)           5c)  5d)      5e)  5f)            5g)  6) r = 3.6 and 0 = 33.7o Answer is: 3.6cis 33.7o        7) a = 5cos 320o and b = 5 sin 320o Answer is: 3.8 - 3.2i       8) z1 = 5.8cis 59o and z2 = 7.8cis 50.2o a) z1z2 = 45.2cis 109.2o b) a = 45.2cos109.2o and b = 45.2sin 109.2o Answer is: -14.9 + 42.7i        9) 9cis 120o 10a) z = 1.4cis 45o and z6 = (1.4)6 cis (6)(45o) = 7.5 cis 270o b) a = 7.5cos 270o and b = 7.5sin 270o Answer is:  -7.5i        11) The three cubed roots of one are found as follows: for k = 0,      = 1cis 120o = -1/2 + .87i for k = 1       = 1cis 240o = -1/2 - .87i for k = 2       = 1cis 360o = 1        This page hosted by  Get your own Free Home Page      I did awesome on this test, take me to the next chapter so I can get a jump on the competition!!    Chapter 13     I didn't do well on this test.  Let me go back and restudy!    Back For free space for your web page --->   Geocities _1156521131.unknown _1156521135.unknown _1156521137.unknown _1156521138.unknown _1156521136.unknown _1156521133.unknown _1156521134.unknown _1156521132.unknown _1156521122.unknown _1156521126.unknown _1156521128.unknown _1156521130.unknown _1156521127.unknown _1156521124.unknown _1156521125.unknown _1156521123.unknown _1156521117.unknown _1156521120.unknown _1156521121.unknown _1156521119.unknown _1156521113.unknown _1156521115.unknown _1156521116.unknown _1156521114.unknown _1156521111.unknown _1156521112.unknown _1156521109.unknown _1156521110.unknown _1156521108.unknown _1156521105.unknown