Results 1 to 6 of 6

Math Help - Complex Numbers Question

  1. #1
    Newbie evanator's Avatar
    Joined
    Mar 2010
    From
    Ireland
    Posts
    24

    Question Complex Numbers Question

    Hey everybody,

    So I have been doing a few exercises on complex numbers, but there is one I can't seem to get out.

    If x and y are real, solve the equation:

    \frac{jx}{1+jy}=\frac{3x+j4}{x+3y}

    I started manipulating it to make it more manageable:

    \frac{jx(x+3y)}{1+jy}=3x+j4

    jx^2+j3xy=(3x+j4)(1+jy)

    etc.

    I have the feeling I'm going nowhere with it. Could somebody please give me a hand?

    Cheers,

    Evanator
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by evanator View Post
    Hey everybody,

    So I have been doing a few exercises on complex numbers, but there is one I can't seem to get out.

    If x and y are real, solve the equation:

    \frac{jx}{1+jy}=\frac{3x+j4}{x+3y}

    I started manipulating it to make it more manageable:

    \frac{jx(x+3y)}{1+jy}=3x+j4

    jx^2+j3xy=(3x+j4)(1+jy)

    etc.

    I have the feeling I'm going nowhere with it.
    In fact, you,re going in exactly the right direction. Next step is to multiply out the brackets on the right (remembering that j^2=-1). Then compare the real parts and the imaginary parts on both sides of the equation.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    \frac{jx}{1+jy}=\frac{3x+j4}{x+3y}<br />

    {(jx)}{(x+3y)}=(1+jy)(3x+j4)

    jx^2+3jxy=3x+4j+3jxy-4y

    jx^2=3x+4j-4y

    so:

    (1) x^2=4

    (2) 0=3x-4y

    Solve this system for x and y
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie evanator's Avatar
    Joined
    Mar 2010
    From
    Ireland
    Posts
    24
    Thank you both of you. Finishing it off:

    jx^2+j3xy=3x+j3xy+j4+(j^2)4y

    0+jx^2=(3x-4y)+j(4)

    x^2=4

    3x-4y=0

    x=\pm2

    Taking the positive value of x and solving the simultaneous equations:

    3x=6

    -3x+4y=0

    4y=6

    y=\frac{6}{4}=1.5

    If we take x=-2 we can see that y will come out as -1.5, so the final results are:

    x=\pm2

    y=\pm1.5
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,543
    Thanks
    1394
    No, that is NOT the answer because x and y are not independent.

    The correct answer is "x= 2 and y= 1.5 or x= -2 and y= -1.5"

    Writing " x= \pm 2, y= \pm 1.5" implies that any combination of those, such as x= 2, y= -1.5, will satisfy the equation and that is not true.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie evanator's Avatar
    Joined
    Mar 2010
    From
    Ireland
    Posts
    24
    Right. Thanks, HallsofIvy. I did understand that the x-value and y-value were dependent on each other, but I didn't realize I was implying otherwise with the plus-minus sign.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. complex numbers question
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: February 28th 2010, 08:10 AM
  2. complex numbers question
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: February 27th 2010, 12:09 AM
  3. Question(s) about complex numbers.
    Posted in the Algebra Forum
    Replies: 5
    Last Post: January 1st 2010, 01:16 PM
  4. Complex numbers question
    Posted in the Pre-Calculus Forum
    Replies: 10
    Last Post: November 3rd 2008, 12:35 PM
  5. complex numbers, a question
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: April 16th 2007, 10:05 AM

Search Tags


/mathhelpforum @mathhelpforum