Results 1 to 10 of 10

Math Help - Complex numbers and square root property

  1. #1
    Member
    Joined
    Apr 2008
    Posts
    79

    Complex numbers and square root property

    Im doing test corrections and need some help. My answers are wrong. Could you please show your work? Thanks.

    simpify

    square root of -12
    I took the i out. I see that 3 and 4 go into 12. I reduced the 4 to 2. I got 3i square root of 2

    square root of -16 and the square root of -49
    Took the i out and got 4. Took the i out and got 7. My answer was 2i square root of 2 i square root of 7.

    Divide the complex numbers
    4 - 3i / 2 - 7i
    I added 2 + 7i on the top. I foiled the problem. and got 8+28i - 6i - 21i^2. My answer was 29 - 22i

    Solve the following equations by using the square root property.

    4x^2 + 16 = 0

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by rowdy3 View Post
    Im doing test corrections and need some help. My answers are wrong. Could you please show your work? Thanks.

    simpify

    square root of -12
    I took the i out. I see that 3 and 4 go into 12. I reduced the 4 to 2. I got 3i square root of 2

    square root of -16 and the square root of -49
    Took the i out and got 4. Took the i out and got 7. My answer was 2i square root of 2 i square root of 7.

    Divide the complex numbers
    4 - 3i / 2 - 7i
    I added 2 + 7i on the top. I foiled the problem. and got 8+28i - 6i - 21i^2. My answer was 29 - 22i

    Solve the following equations by using the square root property.

    4x^2 + 16 = 0

    Thanks.
    if a>0
    \sqrt{-a}=i\sqrt{a}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Apr 2008
    Posts
    79
    These are my answers.

    √(-12)
    = √(12i^2)
    = i√12
    = i(√3 x √4)
    = i(√3 x 2)
    = 2i √3

    √(-16)
    = √(16i^2)
    = i√16
    = 4i

    √(-49)
    = i√49
    = 7i

    (4 - 3i) / (2 - 7i)
    = (4 - 3i)(2 + 7i) / (2 - 7i)(2 + 7i)
    = (8 + 28i - 6i - 21i^2) / (4 - 49i^2)
    = (8 + 22i + 21) / (4 + 49)
    = (29 + 22i) / 53
    = 29/53 + (22/53)i

    4x^2 + 16 = 0
    4(x^2 + 4) = 0
    x^2 + 4 = 0
    x^2 = -4
    x = √-4
    x = √(4i^2)
    x = 2i
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by rowdy3 View Post
    These are my answers.

    √(-12)
    = √(12i^2)
    = i√12
    = i(√3 x √4)
    = i(√3 x 2)
    = 2i √3

    √(-16)
    = √(16i^2)
    = i√16
    = 4i

    √(-49)
    = i√49
    = 7i

    (4 - 3i) / (2 - 7i)
    = (4 - 3i)(2 + 7i) / (2 - 7i)(2 + 7i)
    = (8 + 28i - 6i - 21i^2) / (4 - 49i^2)
    = (8 + 22i + 21) / (4 + 49)
    = (29 + 22i) / 53
    = 29/53 + (22/53)i

    4x^2 + 16 = 0
    4(x^2 + 4) = 0
    x^2 + 4 = 0
    x^2 = -4
    x = √-4
    x = √(4i^2)
    x = 2i
    Good job!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Apr 2008
    Posts
    79
    Which of the following is not equal to i
    a. i^9
    b. i^21
    c i ^33
    d. i^51

    I picked B.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,408
    Knowing that:
    i^{1} = i
    i^{2} = - 1
    i^{3} = i^{2}i = -i
    i^{4} = i^{2}i^{2} = 1
    i^{5} = i^{4}i = i

    The pattern repeats for powers 1 - 4. So:
    If the exponent is 21, then 21/4 = 5 remainder 1 (i.e. go through 5 cycles plus 1 increment). This gives us: i^{21} = i^{1} = i

    So it isn't B but perhaps you can use the same method to determine which of the answers do not equal to i.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Apr 2008
    Posts
    65
    Quote Originally Posted by rowdy3 View Post
    These are my answers.

    √(-12)
    = √(12i^2)
    = i√12
    = i(√3 x √4)
    = i(√3 x 2)
    = 2i √3

    √(-16)
    = √(16i^2)
    = i√16
    = 4i

    √(-49)
    = i√49
    = 7i

    (4 - 3i) / (2 - 7i)
    = (4 - 3i)(2 + 7i) / (2 - 7i)(2 + 7i)
    = (8 + 28i - 6i - 21i^2) / (4 - 49i^2)
    = (8 + 22i + 21) / (4 + 49)
    = (29 + 22i) / 53
    = 29/53 + (22/53)i

    4x^2 + 16 = 0
    4(x^2 + 4) = 0
    x^2 + 4 = 0
    x^2 = -4
    x = √-4
    x = √(4i^2)
    x = 2i
    should not the last one be plus or minus 2i?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Apr 2008
    Posts
    79
    It would be c.

    Solve the equation by completing the quare and applying the square root property.

    w^2 - 16w = 3
    Take half of b and square it. b = -16. -16/2= 8. 8^2=64.
    Add 64 to both sides.
    w^2- 16w + 64= 67
    Factor the Left side
    (w-8)^2= 67
    Using the Square root property take the square root of both sides.
    My answer would be
    w-8 or square root of 67
    w= 8 or the square root of 67
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Apr 2008
    Posts
    79
    Complex numbers
    (2+4i) - (6 - 2i) + (3 - 5i)
    I got - 4 -10i

    Solve the equation by using the square root property

    x^2 - 30 = 0
    Do you divide 30 by 2 and times the 2?

    (y-6)^2 + 4 = 0

    Thanks
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Apr 2008
    Posts
    79
    (2+4i) - (6 - 2i) + (3 - 5i)
    2-6+3 = -1
    4+2-5 = 1
    -1 + i

    x^2 - 30 = 0
    x^2 = 30
    x = +/-sqrt(30)

    (y-6)^2 + 4 = 0
    (y-6)^2 = -4
    y-6 = +/- 2i
    y = 6 + 2i or 6 - 2i
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Complex Analysis Square Root of Z^2
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: September 1st 2011, 12:14 PM
  2. Replies: 3
    Last Post: December 3rd 2009, 09:53 AM
  3. Replies: 0
    Last Post: June 15th 2008, 10:27 PM
  4. use the square root property
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 11th 2007, 02:14 PM
  5. Square Root of Sum of the Numbers
    Posted in the Algebra Forum
    Replies: 5
    Last Post: May 28th 2007, 01:18 PM

Search Tags


/mathhelpforum @mathhelpforum