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Tried getting a common denominator for both fractions

  1. Nov 24, 2004 #1
    find real and complex part of z: z/z+2=2-i

    I cant factor out the z because of the 2 in the denominator. The i can be written as the square root of -1 but that doesnt help. I tried multiplying by the conjugate to get z alone but nope not any good. Im doing something wrong can someone plz start me off ?

    Right now i tried getting a common denominator for both fractions (2z) and then I got stuck at 3z/2z=2-i I think this is wrong.


    Help Me Please :confused:
     
    Last edited: Nov 24, 2004
  2. jcsd
  3. Nov 24, 2004 #2
    Try multiplying both sides by (z + 2)
     
  4. Nov 24, 2004 #3

    Gokul43201

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    Let z = a + ib, cross-multiply and compare real and imaginary parts.
     
  5. Nov 24, 2004 #4
    what am i cross multiplying ?

    I tried multiplying both sides by (z+2) that does not work because then there will be two variable i and z on the right hand side.
     
  6. Nov 24, 2004 #5

    Gokul43201

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    'i' is not a variable. As, you mentioned before, it is the square root of -1. Do you understand what real and imaginary parts mean ?

    If you have an equation of the form a + ib = c + id, then a = c and b = d.
     
  7. Nov 24, 2004 #6
    yes I understand. Well this is what I did, l multiplied both sides by (z+2) this gave me z alone on the left hand side and on the right I got z=(2-i)(z+2) I expanded this and got
    z=2z+4-zi-2i now i dont know what to do

    The solution is z=3+i
     
  8. Nov 24, 2004 #7
    Try to collect all terms "containing" z onto one size of the equation, and factoring.
     
  9. Nov 24, 2004 #8

    Hurkyl

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    You seem to be having a little trouble both with algebra and complex numbers... it might help to try working on similar problems that only involve real numbers. For instance, solve x/(x+2) = 2 for x.
     
  10. Nov 24, 2004 #9
    That I can do x=2x+4 but this question is just hurting my head :cry:
     
  11. Nov 24, 2004 #10

    Gokul43201

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    No...that's not a solution. A solution would look like "x = blah", where 'blah' is a number that does not involve x. Can you get that from x =2x + 4 ?
     
  12. Nov 25, 2004 #11

    BobG

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    Did you try nolachrymose's suggestion. Once you have all the z's on the same side of the equation, you can factor out the z, leaving you with

    z(some numbers)

    Now, if you take and divide both sides of the equation by (some numbers), you can solve the right hand side and do wind up with 3+i, as you said.
     
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