Results 1 to 5 of 5

Math Help - in the set of complex #

  1. #1
    Member
    Joined
    Oct 2006
    Posts
    78

    in the set of complex #

    A- 1) Solve the equation Z^2 5( 1+ i)z + 2+14i = 0

    plz help me??
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Apr 2006
    Posts
    401
    Quote Originally Posted by iceman1 View Post
    A- 1) Solve the equation Z^2 5( 1+ i)z + 2+14i = 0

    plz help me??
    Is "Z" and "z" two different variables? What are you solving for? I'll assume they're the same variable and that you're wanting to solve for z.

    z^2 5*(1 + i)*z + 2 + 14*i = 0

    There is no such z that will satisfy that equation.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Apr 2006
    Posts
    401
    Disregard my previous post, I realized I was using the wrong data.

    There are two solutions that will work for a z, such that that equation is equal to zero.

    Try: 3 + i and 2 + 4*i

    Notice that the simplified equation is:

    z^2 - 5*z - 5*i*z + 2 + 14*i
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Apr 2006
    Posts
    401
    Since,

    z^2 - 5*z - 5*i*z + 2 + 14*i = 0,

    Factor it to get:

    (z - 2 - 4*i)*(z - 3 - i) = 0

    Then:

    (z-2-4*i) = 0 AND
    (z - 3 - i) = 0

    For the first, z = 2 + 4*i
    Second: z = 3 + i
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,842
    Thanks
    320
    Awards
    1
    Quote Originally Posted by AfterShock View Post
    Is "Z" and "z" two different variables? What are you solving for? I'll assume they're the same variable and that you're wanting to solve for z.

    z^2 5*(1 + i)*z + 2 + 14*i = 0

    There is no such z that will satisfy that equation.
    AfterShock's factoring method is better than this one, but remember any quadratic with real or complex coefficients may be solved using the quadratic formula. The solution in this case is a bit of a mess (which is why I recommend factoring whenever possible) but will work.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: November 27th 2011, 12:15 AM
  2. Replies: 3
    Last Post: October 4th 2011, 05:30 AM
  3. Replies: 6
    Last Post: September 13th 2011, 07:16 AM
  4. Replies: 12
    Last Post: June 2nd 2010, 02:30 PM
  5. Replies: 1
    Last Post: March 3rd 2008, 07:17 AM

Search Tags


/mathhelpforum @mathhelpforum