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a + jb

Discussion in 'Homework Help' started by shermaine, May 5, 2009.

  1. shermaine

    shermaine New Member

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    Hi,

    How do i solve this equation of (z - 1)^4 = (1 + J)^10 / (1-J) in form of (a + jb)?
     
  2. dknguyen

    dknguyen Well-Known Member

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    You would solve it like any other equation...isolate z. Then you manipulate the expression to get it into the form of a+jb. One trick is that since j*j = -1, you can multiply expressions like (x+jy) by it's conjugate divided by it's conjugateTHis means you would multiply the expression by (x-jy) / (x-jy) = 1. Since it's equal to one, you aren't changing the expression. YOu are just changing the way it looks.

    (x+jy)*(x-jy)/(x-jy) = (x^2+y^2)/(x-jy).

    What this lets you do is move the complex j expression between the numerator and denominator. ANd if you do it right, it probably divides itself out, or puts the j in a place where you want it.
     
    Last edited: May 5, 2009
  3. Tesla23

    Tesla23 Member

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    Try converting the RHS to polar form and taking the 4th root.

    once you have found one solution, try to find the other three.
     
  4. dave

    Dave New Member

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  5. Miles Prower

    Miles Prower Member

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    First thing I would do is convert that mess on the right side of the '=' sign to Steinmetz form.

    If: C= a + jb (Cartesian) then

    R= sqrt(a^2 + b^2)

    Theta= arctan(b/a)

    C= R /_ Theta (Steinmetz)

    That makes exponentiation and division relatively straight forward.

    C^x= R^x /_ (Theta * x)

    Division:

    A= Ra /_ Thetaa

    B= Rb /_ Thetab

    A / B= (Ra / Rb) /_ (Thetaa - Thetab)

    Multiplication:

    A X B= (Ra X Rb) /_ (Thetaa + Thetab)

    If you want it in Cartesian form:

    a= Rcos(Theta)

    b= jRsin(Theta)
     
    Last edited: May 5, 2009

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