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A+bi form, complex analysis

  1. Aug 27, 2011 #1
    1. The problem statement, all variables and given/known data

    Use the Definition Re(z1)=Re(z2), Im(z1)=Im(z2)to solve each equation for z=a+bi.

    [itex]\frac{z}{1+\bar{z}}[/itex]=3+4i



    2. Relevant equations

    Sec 1.1 #42 from Complex Analysis 2nd ed from Dennis Zill

    3. The attempt at a solution
    I have solved several similar problems like this one in my text but I'm getting stuck on this one part.

    The goal is to say:

    [itex]\frac{z}{1+\bar{z}}[/itex]=[itex]\frac{a+ib}{1+a-ib}[/itex]


    and put the right hand side of this equation into a real part and an imaginary part and equate the real and imaginary parts to the original one given.

    So in short how to I put [itex]\frac{a+ib}{1+a-ib}[/itex] into a+bi form?

    I have tried many conjugates but none have worked

    Thanks
     
  2. jcsd
  3. Aug 27, 2011 #2

    Char. Limit

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    Gold Member

    It might help if you carried out the division directly, i.e. remember that:

    [tex]\frac{a+ib}{c+id} = \frac{ac+bd}{c^2+d^2} + i \frac{bc - ad}{c^2 + d^2}[/tex]

    Just let a+1=c and b=d for your division.
     
  4. Aug 27, 2011 #3

    ehild

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    That is not needed. Just substitute z=a+ib for z in the original equation, multiply both sides with the denominator, and compare both the real and imaginary parts on each sides.

    ehild
     
  5. Aug 27, 2011 #4
    Thank you ehild, I got it with your advice. For some reason I just forgot that the real and imaginary parts can have both a's and b's in it.
     
  6. Aug 27, 2011 #5

    HallsofIvy

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    Another way to solve
    [tex]\frac{a+ ib}{1+a- ib}= 3+ 4i[/tex]
    is to multiply both sides by 1+ a- ib:
    a+ ib= (3+ 4i)(1+ a- ib).

    Multiply the right side out and equate real and imaginary parts.
     
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