Friday, March 6, 2009

Complex Numbers Set 4

  1. x = 3 - 2j and y = 3 + 2j

    Compute:
    x + y
    x - y
    x^2
    y^2
    xy
    (x + y)(x - y)
  2. Write the complex number in standard form (i.e. in the form a + b i.)
    1. 3+ √(-16)
    2. -5j + 3 j^2
  3. Evaluate (2 + 3j) - (6 -4j)

b) (2 + 3j)(1 - 5j)

c) 7j(7 - 3i)

d) (3 + 2j)^2 + (4 - 3j)

4. Write the conjugate of the complex number.
a) 3 + 5j
b) 2 - 6j
c) 15i
5. Perform the operation and write the result in standard form.
a) 4/(2+3j)
b) (3 + 4j)/(3-2j)
c) 3/(3 + 2j) - 4/(3 -2j)
V. Solve the quadratic equations using the Quadratic Formula.
a) 3x^2 + 9x +7 = 0
b) y^2 -2y + 2 = 0

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43 comments:

  1. Desi GirlMarch 6, 2009 at 2:44 PM

    x = 3 - 2j and y = 3 + 2j

    Compute:
    x + y
    wel substutite values for x and y..
    (3-2j)+(3+2j)
    Place the real under the real and imaginary under imaginary.The real nos. are 3 and 3 and the imaginary nos. are -2j and +2j so we get 9+0j.

    ReplyDelete
  2. Desi GirlMarch 6, 2009 at 2:48 PM

    Write the conjugate of the complex number.
    a) 3 + 5j
    b) 2 - 6j
    c) 15i

    If am not mistaken the conjugate for (A)3-5j
    (b) 2+6j AND (c)-15i
    Its jus the opposite, correct if am wrong plz

    ReplyDelete
  3. Dark AngelMarch 6, 2009 at 4:35 PM

    Hey Desi Girl...

    you are correct for (a) and (b) but im not sure about (c).

    can someone explain...??

    ReplyDelete
  4. ridiclyricMarch 7, 2009 at 11:03 AM

    I'm not too sure of part c Desi Girl but i can say that any complex number multiplied by its conjugate is suppose to give a real number....

    ReplyDelete
  5. ridiclyricMarch 7, 2009 at 11:09 AM

    .......What I said before implies that you are correct because

    15i × -15i = -15i^2

    Recall that i^2 = -1

    Therefore -15i^2= -15 (-1)

    Final answer is 15

    Therefore -15i is the conjugate of 15i

    Hope that I have made it clear for both desi girl and dark angel

    ReplyDelete
  6. poopsiepumkinpieMarch 9, 2009 at 6:36 PM

    c) 7j(7 - 3i)

    49j- 21j^2
    because i and j is d same ting!!!

    49j + 21

    ReplyDelete
  7. mysticwings*March 9, 2009 at 10:25 PM

    1) x+y = (3-2j) + (3+2j)
    = 6

    x-y = (3-2j) - (3+2j)
    = -4j

    x^2 = (3-2j)^2
    = (3-2j)(3-2j)
    = 9-6j-6j+4j^2
    = 9-12j+4(-1)
    = 5-12j

    y^2 = (3+2j)^2
    = (3+2j)(3+2j)
    = 9+6j+6j+4j^2
    = 9+12j+4(-1)
    = 5+12j

    xy = (3-2j)(3+2j)
    = 9 - 4j^2
    = 9 +4
    = 13


    (x+y)x(x-y)-
    =(3-2j)+(3+2j)x(3-2j)-(3+2j)
    = 9 x -4j
    =-36j

    ReplyDelete
  8. poopsiepumkinpieMarch 10, 2009 at 6:50 AM

    3/(3 + 2j) - 4/(3 -2j)

    in this ques u wud work out the 1st part 3/(3 + 2j) and the conjugate would be 3-2j.

    when u multiply out everything den u do the same forthe second part 4/(3 -2j) where the conjugate would be 3+2j. when u obtain both ans u jus minus them to get your final ans.

    ReplyDelete
  9. poopsiepumkinpieMarch 10, 2009 at 6:52 AM

    3+ √(-16)
    3+ √16j
    which is 3+4j

    ReplyDelete
  10. mysticwings*March 10, 2009 at 9:34 AM

    no 2
    a)3+(-16)^1/2 = 3 + (-1 x 16)^1/2
    = 3 + 4j

    b)-5j+3j^2= -5j + 3(-1)
    = -3 - 5j

    ReplyDelete
  11. mysticwings*March 10, 2009 at 10:02 AM

    no 3
    a) (2+3j) - (6-4j)= -4 +7j

    b)(2+3j)(1-5j)= 2-10j+3j-15j^2
    = 2-7j-15(-1)
    = 17-7j

    c)7j(7-3i)= 49j-21j^2
    = 49j-21(-1)
    = 21+49j

    d)(3+2j)^2 + (4-3j)= (3+2j)(3+2j)+(4-3j)
    = (9+12j+4(-1))+ (4-3j)
    = (5+12j)+(4-3j)
    = 9+9j

    ReplyDelete
  12. mysticwings*March 10, 2009 at 10:20 AM

    no 4
    the cojugates are
    a)3-5j
    b)2+6j
    c)-15i
    NB all that changes is the opperation sign.
    positive changes to negative and vice versa

    ReplyDelete
  13. empressMarch 10, 2009 at 12:28 PM

    mysticwings, remember the saying of BODMAS(brackets of division multiplication addition then subtraction) try and apply it plz. no offense.
    for: (x+y)x(x-y)please do brackets FIRST
    remember you worked out x+y to be 6 and x-y to be -4j: then, you could of just used it to get:
    (6)(-4j)= -24j

    ReplyDelete
  14. empressMarch 10, 2009 at 12:33 PM

    hey people:
    i means the same as j so just calculate as usual

    ReplyDelete
  15. empressMarch 10, 2009 at 12:36 PM

    therfore c will be:
    7j(7-3i)= 7j(7-3j)
    = 49j-21j^2
    = 49j-21(-1){NB. j^2=-1}
    = 49j+21

    ReplyDelete
  16. R@ptorMarch 11, 2009 at 6:19 PM

    x = 3 - 2j and y = 3 + 2j
    therefore:
    x+y = (3-2j) + (3+2j)
    = 6

    x-y = (3-2j) - (3+2j)
    = -4j

    x^2 = (3-2j)^2
    = (3-2j)*(3-2j)
    = 9-6j-6j+4j^2
    = 9-12j+4(-1)
    = 5-12j

    y^2 = (3+2j)^2
    = (3 + 2j)(3 + 2j)
    = 9 + 6j + 6j + 4j^2
    = 9 + 12j + 4(-1)
    = 5 + 12j

    xy = (3-2j) (3+2j)
    = 9 - 4j^2
    = 9 +4
    = 13


    (x+y)*(x-y)-
    =(3-2j) + (3+ 2j) x (3-2j) -(3+2j)
    = 9 x -4j
    =-36j

    ReplyDelete
  17. the best oneMarch 12, 2009 at 9:37 AM

    no 2
    a)3+(-16)^1/2 = 3 + (-1 x 16)^1/2
    = 3 + 4j

    b)-5j+3j^2= -5j + 3(-1)
    = -3 - 5j

    ReplyDelete
  18. santoshMarch 12, 2009 at 11:49 AM

    The conjugate is the opposite sign

    The conjugate of 3 + 5j is 3 – 5j
    The conjugate of 2 – 6j is 2 + 6j
    The conjugate of 15i is -15i

    ReplyDelete
  19. gunironMarch 12, 2009 at 11:58 AM

    QUESTION 2

    3-√(-16)

    Firstly, the root of a negative number gives an error so that √(-16) must be expressed with its imaginary part. √(-16 can be split into √(16) * √(-1)

    Therefore: 3 - √(-16) = 3 - √16 * √-1 but: √-1 = j
    = 3 - √(16)j
    Now we can determine the √16
    = 3 – 4j

    ReplyDelete
  20. gunironMarch 12, 2009 at 12:03 PM

    Question 2b

    -5j + 3j2
    To evaluate this question we must no that j2 = -1, so that it may be substituted into the equation
    Therefore: -5j + 3j2 = -5j + (3 * -1)
    = -5j – 3
    = -3 – 5j

    ReplyDelete
  21. boy-bMarch 12, 2009 at 1:40 PM

    15i=15 x i so the opposite of that is 15 divided by i..........this dont make no sense!

    ReplyDelete
  22. gunironMarch 12, 2009 at 3:10 PM

    SET 04 QUESTION 01

    x = 3-2j and y =3+2j

    1) for x+y we substitute x=3-2j and y=3+2j into the equation
    x+y = (3-2j) + (3+2j)
    group real parts with real parts and imaginary parts with imaginary parts

    3 - 2j
    + 3 + 2j
    = 6 + 0j


    2) x - y
    The same approach is used for subtraction, i.e. substitute and group the like terms
    x - y = (3 - 2j) - (3 + 2j)

    3 - 2j
    - 3 + 2j
    = 0 - 4j



    3) x^2
    Substitute x =(3 - 2j)
    therefore (3 - 2j)^2
    =(3 - 2j)* (3 - 2j)
    Using the loop process let us evaluate
    (a + b) * (c + d)= ac + ad + bc + bd
    therefore (3 - 2j)*(3-2j)= 9 - 6j -6j + 4j^2
    = 9 - 12j + 4j^2
    but: j^2 = -1 = 9 - 12j + 4*(-1)
    = 9 - 12j - 4
    = 9 -4 - 12j
    = 5 - 12j

    4) y^2 uses the same concepy of the loop
    y = (3 + 2j)
    (3 + 2j) * (3 + 2j)
    = 9 + 6j +6j + 4j^2 but: j^2 = -1
    = 9 + 12j + 4*(-1)
    = 9 + 12j -4
    = 9 - 4 + 12j
    = 5 + 12j

    ReplyDelete
  23. gunironMarch 12, 2009 at 3:19 PM

    PART 5 QUESTION 01
    xy substitute x=(3 - 2j) and y=(3 + 2j)

    therefore: (3 - 2j) * (3 + 2j)

    this is known as the difference of squares
    that is, (a - b)* (a + b) = (a^2) - (b^2)

    so that: a = 3 and b = 2j

    = (3^2) - (2j)^2
    = 9 - 4j^2 but: j^2= -1
    = 9 - (4*-1)
    = 9 - (-4)
    = 9 + 4
    = 13
    for complex numbers the difference of squares when expanded and simplified should give a real number only

    ReplyDelete
  24. gunironMarch 12, 2009 at 3:23 PM

    for part 06 QUESTION 01
    (x + y )*(x - y) always evaluate the brackets first
    from part 01, x+y = 6 +0j = 6
    from part 02, x -y= 0 - 4j = -4j

    therefore: 6 * -4j = -24j
    atleast i think so

    ReplyDelete
  25. mysticwings*March 12, 2009 at 5:41 PM

    empress, i dont know what stupidness you are talking about because i did follow BODMAS, i jus made a mistake and multiply 3 by 3 instead of adding it!!!!
    my calculation may have slipped but the way i work out problems is correct!!!!!

    ReplyDelete
  26. boy-bMarch 14, 2009 at 11:23 AM

    empress u was talkin bout using BODMAS for this concept (x+y)x(x-y), i dont think it wod matter
    eg let x=2 y=1
    proof1: (x+y)x(x-y)
    (2+1)x(2-1)
    3 x 1 = 3
    proof2: (x+y)x(x-y)
    (2+1)x(2-1)
    4-2+2-1
    4-1
    =3

    ReplyDelete
  27. fariel'sbeststudentMarch 14, 2009 at 11:33 AM

    first of all guniron, you didn't have to go to all such lengths as to get the result

    you simply could have said that

    6 multiplied by the -4j

    can be written as .................

    ummmmm okay i now get where i was going and i think it wrong,

    i was thinking that we could multiply the 6 by the -4j as if two similar factors like (6 * 4)j

    correct me plz

    ReplyDelete
  28. BladeMarch 14, 2009 at 9:31 PM

    yeah guniron for part qu1. part 6 that sounds correct....working out inside the brackets and then multiplying.

    For the last set is the quadratic formular;
    x=[b+/-(√b^2 -4ac)]/2a ?

    ReplyDelete
  29. SexyloveMarch 16, 2009 at 9:58 AM

    The complex * by the conjugate suppose to give the real number and the conjugate is the opposite sign..dat is wat i know

    ReplyDelete
  30. damonMarch 16, 2009 at 10:32 AM

    x = 3 - 2j and y = 3 + 2j

    x + y = (3 - 2j) + (3 + 2j)
    = (6 + 0j)
    = 6

    ReplyDelete
  31. damonMarch 16, 2009 at 10:35 AM

    a) 3 + 5j
    =(3 - 5j)

    b) 2 - 6j
    =(2 + 6j)

    c) 15i
    = -15i

    ReplyDelete
  32. UnknownMarch 16, 2009 at 11:05 AM

    2d) (3+2j)^2 + (4-3j)
    (3+2j)^2 = (3+2j)(3+2j) = 5+12j
    (5+12j)+(4-3j) = 9+6j

    ReplyDelete
  33. It's meMarch 16, 2009 at 2:12 PM

    Question 3

    a) (2+3j)-(6-4j)
    add real no. with real no. and imaginary no. with imaginary no.
    therefore real no. 2-6 = -4
    imaginary no. 3j-(-4j) =7j
    so the answer is -4+7j

    b) (2+3j)(1-5j)
    firstly u expand the brackets so therefore it will be first term from the first bracket by each term in the second bracket and then second term from the first bracket by the both term from second brackets and u get the following
    2*1+2*-5j+3j*1+3j*-5j
    = 2+(-10j)+3j+(-15j^2)
    from there u now group all real no. with real no.and imaginary no. with imaginary no.
    = 2+(-10j)+3j-15j^2
    = 2+(-7j)-15j^2
    remember j^2 = -1 so therefore
    2+(-7j)(-15*-1)
    2+(-7j)+15
    add now add two real numbers
    2+15-7j
    answer = 17-7j

    c) 7j(7-3i)
    expand the equation so therefore
    7j*7-7*3i
    = 49j-21j^2
    all way remember j^2 =-1
    so 49j(-21*-1)
    = 49j=21

    d)(3+2j)^2 +(4-3j)
    firstly you have to expand the first brackets because it squared so therefore
    (3+2j)(3+2j)
    first term from the first bracket by each term in the second bracket and then second term from the first bracket by the both term from second brackets and u get the following
    (3*3+3*2j+2j*3+2j*2j)
    = 9+6j+6j+4j^2
    = 9+12j+4j^2
    remember j^2 =-1
    therefore 9+12j+(4*-1)
    = 9+12j-4
    =5+12j
    now you add this answer to the second part of the question therfore
    (5+12j)+(4-3j)
    add real no. with real no. and imaginary no. with imaginary no
    = 5+4+12j-3j
    answer = 9+9j

    ReplyDelete
  34. DarkyMarch 20, 2009 at 5:46 PM

    x = 3 - 2j   y = 3 + 2jx + y (3 - 2j) + (3 + 2j) 6 Desigirl I am not clear on what you did in your first comment is that not multipying when you should be adding

    ReplyDelete
  35. DarkyMarch 20, 2009 at 5:50 PM

    ( 2 + 3 j ) - ( 6-4j)
    remember your real and imaginery

    -4 -j

    ReplyDelete
  36. Kurosaki IchigoMarch 21, 2009 at 8:05 AM

    I’ll just like to add a comment about the conjugate. I think that the purpose of conjugating a base with respect to complex numbers is to multiply it by its original to get rid a the denominator.

    ReplyDelete
  37. Low riderApril 2, 2009 at 6:08 PM

    I think that is the purpose of the conjugate. Actually i don't think there is any other way to solve problems like those.

    ReplyDelete
  38. turboApril 11, 2009 at 4:56 PM

    yes Low rider and kurosaki lchigo, it is correct...well as far as i know

    ReplyDelete
  39. Chun-LiApril 16, 2009 at 8:18 PM

    yeah dat's true, multiplying the complex number by the conjugate simplifies division of complex numbers, since it changes the denominator from imaginary to a real number

    ReplyDelete
  40. Chun-LiApril 16, 2009 at 8:53 PM

    Question 3c ........7j ( 7- 3i )
    i think dat 'poopsiepumpkinpie' made a lil error...instead of d ans. being 49j + 21
    its 49j - 21
    this is so because when u calculate 7j ( 7-3i ), i and j is d same ting u'l get...... 49j - 21j^2

    which gives, 49j - 21 (-1 x -1)

    so the ans. is .....49j-21

    ReplyDelete
  41. Chun-LiApril 16, 2009 at 9:13 PM

    oh no wait...i'm so sorry my bad...lol its
    49j + 21 since j^2 = -1...for a sec i got mixed up n thought j= -1

    so it'l b 49j - 21 (-1) hence, 49j + 21

    like miss always says...look out 4 police lol...be careful

    ReplyDelete
  42. Chun-LiApril 16, 2009 at 9:17 PM

    Question 3d) (3 + 2j)^2 + (4 - 3j)

    firstly...(3 + 2j)^2 gives
    (3 + 2j)(3 + 2j) = 18 + 6j + 4j^2
    ...... = 18 + 6j + 4(-1)
    ...... = 18 + 6j - 4
    ...... = 14 + 6j
    therefore,
    (14 + 6j) + (4 - 3j)

    = 18 + 3j

    ReplyDelete
  43. Chun-LiApril 16, 2009 at 9:25 PM

    hi guys umm i need help with the last question

    V. Solve the quadratic equations using the Quadratic Formula.
    a) 3x^2 + 9x +7 = 0
    b) y^2 -2y + 2 = 0

    i'm not sure how to start...i don't see the connection with complex numbers..help!

    ReplyDelete

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