Thursday, March 5, 2009

complex numbers questions 2

  1. Evaluate the following: √(-25) + √(64) added to √(81) + √(-9)
  2. Evaluate the following: √(-25) + √(64) divide by √(81) + √(-9)
  3. Evaluate the following: √(-25) + √(64) minus √(81) + √(-9)
  4. Evaluate the following: √(-25) + √(64) multiply by √(81) + √(-9)
  5. Evaluate the following: j^6 + j^7 added to j^3 + j^4

39 comments:

  1. Q1. multiply the root of -1 by the root of 25 and you will get the root of 25j(you will get a positive value as j=the root of -1)and you get 5j and then find the root of 64 which is =8.So the ans for this part is 5j+8.
    For the the next part of the question follow the same concept as the first part and you will get 3j+9.
    The overal answer for this question is will be
    5j+8
    +
    3j+9 = 8j+17

    ReplyDelete
  2. Q2. tapolin already worked out the first two parts of the problem so the final part is:

    (5j+8)/(3j+9)
    to divide complex numbers, you must multiply by the conjugate.
    the question now is:

    [(5j+8)/(3j+9)]*[(3j-9)/(3j-9)]=

    (15j^2+24j-45j-72)/(9j^2+27j-27j-81)=

    (15j^2-21j-72)/(9j^2-81)=

    (-15-72-21j)/(-9-81)=

    (-87-21j)/(-90)

    ReplyDelete
  3. (√-25 + √ 64) + (√ 81 + √ -9)

    Recall that √ (-8) = √ -1× √ 8

    This implies: √ -1× √25 +√ 64 = 5j + 8
    +
    √ -1× √9 + √ 81= 3j + 9

    Ans = 8j + 17

    ReplyDelete
  4. (j^6 + j^7) + ( j^3 + j^4)

    Recall j^2= -1

    So
    j^6 = j^2 × j^2 × j^2
    = -1 × -1 × -1
    = -1

    j^7 = j^2 × j^2 × j^2 × j
    = -1× -1 ×-1 × √-1
    = -j

    j^3 = j^2× j
    = -1 × j
    = -j

    j^4 = j^2 × j^2
    = -1 × -1
    = 1

    this implies that -1 +(-j) +(-j) + 1 = -2j

    ReplyDelete
  5. (√ (-25) + √64) ÷ (√81 + √ (-9))

    This Implies:

    √ (-25) + √ 64
    √ 81 + √ (-9)

    In order to solve this we must first make the denominator a real number by multiplying it by its conjugate. The conjugate is the same the denominator but with opposite signs. (The conjugate must be equal to one when simplified) Therefore in this case the conjugate is

    √ 81 - √ (-9)
    √ 81 - √ (-9)

    Therefore this is what we should have:

    √ (-25) + √ 64 × √ 81 - √ (-9)
    √ 81 + √ (-9) √ 81 - √ (-9)

    It can be seen that the conjugate is equal to one

    Now we can apply the difference of two squares
    (a2 + b2) = (a + b) (a – b)

    Calculation for Denominator:
    Applying difference of two squares:

    This implies:
    (√ 81)2 – (√ -9)2

    Recall √-9 = √-1 ×√9 = 3j

    This implies
    81- 3j^2
    j^2 = -1

    Therefore 81-(-3) = 81+3= 84

    Calculation for Numerator

    (√ (-25) + √64) (√ 81+ √ (-9))

    (√ (-25) × √81) + (√ (-25) × √(-9)) + (√64 × √81) + (√64× √(-9))

    Now Simplify

    (5j × 9) + (5j × 3j) + (8 × 9) + (8 × 3j)

    45j + 15j^2 + 72 + 24j

    Recall j^2 = -1

    Then Group and solve:

    (72 – 15) + (45j + 24j) = 57 + 69j


    Overall Solution

    Answer: 57 + 69j
    84

    ReplyDelete
  6. please note that the ans to the previous ques is (57 + 69j) ÷ 84, sorry the line didn’t come out. ( same with the conjugate)

    ReplyDelete
  7. Q3.√(-25) + √(64)-√(81) + √(-9)
    √(-25) + √(64)
    √-1√25 + 8
    =5j+8
    √(81) + √(-9)
    =9+√-1√9
    =9+3j
    Therefore: 8+5j
    -9+3j
    Ans =-1+2j

    ReplyDelete
  8. hey correct me if im wrong but isnt tapolin wrong about the root of -25??
    the root of -25=root of 25 * root of -1
    therefore it would be root 25j as the final answer?doesnt the root apply over the entire 25j or just the 25?
    in which case is it right to find the root of j?

    ReplyDelete
  9. Yes taplin is correct about root 25 being 5.

    (root 25)* (root-1)
    (the root of 25 is 5) and the (root of -1 is j)
    so simply the answer is 5j. do not get mistaken, (root(25*-1)) is not the same as (root25 * root -1)

    ReplyDelete
  10. hey samos u hav 2find d square root of the 25 first since it is a whole no.Understand??

    ReplyDelete
  11. aye jade and ridiclyric the no. 3 u did am not gettin ne of those.This is wat am getting:

    √ (-25) + √ (64) divide by √ (81) + √ (-9)

    From this u can get:

    √-1 √25 and the √ (64) is 8
    Since the √-1 is j and wel the √25 is 5
    This nw reads : 5j+8


    √ (81) + √ (-9)

    The √ (81) is 9 and the √ (-9 can be broken
    Up into two pieces, √-1 √9.
    The √9 is 3 and wel the √-1 is j

    NOW: 5j + 8/3j + 9 X 3j - 9/3j – 9


    The numerator
    ( 5j +8) (3j- 9)
    15j2 – 45 + 24j – 72
    15j2-21j-72
    15(-1)- 21j – 72
    -15 – 21j-72
    -15-72-21j
    -87-21j

    The denominator
    3j – 9
    (3j) – (9) 2
    9j2 + 81
    9(-1) + 81
    -9+ 81
    72

    1/72 ( 87 – 21j)

    ReplyDelete
  12. desi girl question 3 is not devide its minus, its question 2 you have to devide for

    ReplyDelete
  13. desi girl i kind ah lost when i reach up to the part (NOW: 5j + 8/3j + 9 X 3j - 9/3j – 9) how you get that.

    ReplyDelete
  14. wel angel i worked each part at a time and then brought them 2gether...

    ReplyDelete
  15. for Q2 why can't the answer be worked like this:
    sq.root-25 + sq.root 64 - sq.root81 + sq.root-9
    which will give:
    5j + 8 - 9 + 3j
    5j + 3j - 9 + 8
    = 8j -1
    instead of:
    5j + 8
    -3j -9
    =2j - 1

    ReplyDelete
  16. well desi girl how i see it is the denominator will be the same as the differenc of two squares since they are the same but one has a minus and one has a plus.So i applied the equation a^2 - b^2 = ( a + b ) ( a - b). Then i recalled that the root of a number squared is equal to the number. that is how i came up with that answer. It may be kinda confusing because some of the division and brackets did not copy. Please let me kno if you understood how i came to my answer....in ques 2

    ReplyDelete
  17. i dont understand what ridiclyric did for Q2.
    isn't the answer supposed to be worked out like this?:
    the numerator sq.root-25+sq.root64 divided by the denominator sq.root81+sq.root-9
    TAKING NUMERATOR
    =>sq. root -25 + sq.root 64
    5j + 8
    TAKING DENOMINATOR
    =>sq.root 81 + sq.root -9
    9 + 3j
    THEREFORE:
    (5j + 8/3j + 9) x (3j - 9/3j - 9(conjugate))

    SO NEW NUMERATOR = (5j + 8)x(3j - 9)
    NEW DENOMINATOR = (3j + 9)x(3j - 9)

    taking denominator:
    (3j+9)x(3j-9)
    9j^2-27j+27j-81
    9(-1)-81
    -9-81
    =-90

    taking numerator:
    (5j+8)x(3j-9)
    15j^2-45j+24j-72
    15(-1)-21j-72
    -15-72-21j
    -87-21j

    TOTAL RESULT:
    N/D: -87-21j/90

    ReplyDelete
  18. why do we need to use the denominator particulary to to find the conjugate?
    why cant we use the numerator instead?

    ReplyDelete
  19. Hey guys I realise dat you all don’t understand how I got my answer but I will be pleased if you showed me exactly where I went wrong instead of just sayin you don’t know what I did. However I do see one error in the calculation of my denominator with (√-9) ^2 this is suppose to work out to (√-1)2 (√ 9)^2 = (j^2) (3)^2

    = 9j

    so it is 81 – 9j^2 = 81 – (-9)

    81+9= 90


    Therefore my overall solution should be (57+69j) ÷ 90

    ReplyDelete
  20. Hey I found my problem in my calculation for the numerator I mixed up the signs, so instead of multiplying by the conjugate I multiplied by the original equation. Thanks next time I’ll try to do it a lil slower before I make that stupid error again……thanks

    ReplyDelete
  21. This comment has been removed by the author.

    ReplyDelete
  22. no 1
    the first thing to do is to work out anything that can be worked out! dont trail anything, its makes the problem seem difficult

    = (-1x25)^1/2 + 8 + 9 + (-1x9)^1/2
    =5j + 17 +3j
    17 + 8j

    ReplyDelete
  23. no 2
    the key to this problem is again to work out anything that can be worked out!!! then multiply by the conjugate
    = [(8+5j)/(9+3j)] x [(9-3j)(9-3j)]

    = (72-24j+45j-15j^2)/(81-9j^2)

    =[72-21j-15(-1)]/[81-9(-1)]

    = (87 - 21j)/90

    ReplyDelete
  24. no 3
    work out what can be worked out then take away the terms that are in common. pay special attention to opperation signs!!!

    =(8+5j) - ((9+3j)

    =(-1 + 2j)

    ReplyDelete
  25. no 4
    work out what can be worked out by removing the "shed" ie the square root sign.
    = (8+5j) x (9+3j)
    multiply each term in the first bracket by each term in the second bracket.

    = 72 + 24j + 45j + 15j^2
    = 72 + 69j -15
    = 57 + 69j

    ReplyDelete
  26. no 5

    j^6 + j^7 + j^3 + j^4 = 1

    an easy way to do this is to add all the powers,and since we know that j^2 = -1 we divide the total number of powers by 2. if we get an even number then the answer is 1, but if we get an odd number then the answer is -1.

    ReplyDelete
  27. (Q2) {√ (-25) + √ (64)} ∕ {√ (81) + √ (-9)}
    let’s call √ (-25) + √ (64) _ X
    and √ (81) + √ (-9) _ Y
    to divide X by Y , we have to multiply the numerator, X and the denominator, Y by the conjugate of Y.
    Two complex numbers of the form a+bj and a-bj are said to be conjugate

    Numerator: √ (-25) + √ (64) → (√ -1 * 25) + 8
    = j √ 25 + 8
    = 5j + 8

    Denominator: √ (81) + √ (-9) → 9 + (√ -1 * 9)
    = 9 + j √ 9
    = 9+ 3j

    X/ Y : 5j + 8/9+ 3j____________ the conjugate of Y is 9- 3j

    Numerator (5j + 8) (9- 3j) = 87+ 2j

    Denominator (9+ 3j) (9- 3j) = 90

    Ans. 1/90 (87 +21j)

    ReplyDelete
  28. so the conjugate is simply just the opposite of the denominator. and just multiply the entire equation by its conjugate...? so how come it becomes 1/90 (87-21j) why is it 1/90 and not 90 as found..

    ReplyDelete
  29. Santosh i quite don't understand the ending of the question could you probly show all your steps.

    ReplyDelete
  30. #

    {√(-25) + √(64)} + {√(81) + √(-9)}
    {√25√-1 + √(64)} + {√(81) + √9√-1}
    (5√-1 + 8) + (9 + 3√-1
    using: √-1 = j
    5j + 8 + 9 + 3j
    adreal to real and imag. to imag.
    17 + 8j

    ReplyDelete
  31. # 5

    (j^6 + j^7) + (j^3 + j^4)
    using: j^6 = j^2 * j^2* j^2
    j^7 = j^2 * j^2* j^2 * j
    j^3 = j^2 * j
    j^4 = j^2 * j^2
    considering: j = √-1
    and j^2 = -1
    we get:j^6 = -1 * -1 * -1
    j^7 = -1 * -1 * -1 * √-1
    j^3 = -1 * √-1
    j^4 = -1 * -1

    and: j^6 = -1
    j^7 = -1 * √-1 = -1j
    j^3 = -1 * √-1 = -1j
    j^4 = 1
    therfore we get:(j^6 + j^7) + (j^3 + j^4)
    = -1 + -1j + -1j +1
    = -2j

    ReplyDelete
  32. Question 2
    the key is to simplify the question:
    = [(8+5j)/(9+3j)] x [(9-3j)x(9-3j)]

    = (72-24j+45j-15j^2)/(81-9j^2)

    =[72-21j-15(-1)]/[81-9(-1)]

    = (87 - 21j)/90

    Do anyone diasagree with my answer??

    ReplyDelete
  33. Question 3

    √(-25) + √(64) minus √(81) + √(-9)

    Firstly work out what can be worked out then take away the terms that are in common.


    =(8+5j) - (9+3j)

    Real - Real.........imag - imag
    =(-1 + 2j)

    ReplyDelete
  34. Evaluate the following: √(-25) + √(64) added to √(81) + √(-9)

    keepin in mind that

    √(-25) = √(-1) x √(25) = 5j

    √(-9) = √(-1) x √(9) = 3j

    rewrite and add like terms

    (8+5j) + (9+3j)

    8 + 5j
    +
    9 + 3j
    _______

    17+ 8j
    _______

    .

    ReplyDelete
  35. Evaluate the following: √(-25) + √(64) minus √(81) + √(-9)

    to solve this question follow the steps taken in the previous question i did and then minus like terms

    8 + 5j
    -
    9 + 3j
    _______

    -1+ 2j
    _______

    ReplyDelete
  36. Evaluate the following: √(-25) + √(64) divide by √(81) + √(-9)

    to solve this question follow the steps taken in the previous question i did and then devide


    (8 + 5j)÷(9 + 3j)

    ReplyDelete
  37. the same goes for multiplication

    (8 + 5j)x(9 + 3j)

    ReplyDelete