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Complex Number Problem

  1. Jun 12, 2008 #1
    Hi, i have this question which is related to complex number and i have just no idea how i should solve it. Some guide and help please.

    Given that z = x + yi and w = (z+8i)/(z-6) , z [tex]\neq[/tex] 6. If w is totally imaginary, show that x^2 + y^2 + 2x - 48 = 0

    I've tried alot of way comparing them. Just can't work.
    I substituted z into w but end up still with a w. How can i get rid of the w?
     
    Last edited: Jun 12, 2008
  2. jcsd
  3. Jun 12, 2008 #2

    tiny-tim

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    Hi crays! :smile:
    You should really have shown us what you got, otherwise we can't see where you went wrong.

    Hint: multiply top and bottom by the complex conjugate of (z - 6). :smile:
     
  4. Jun 12, 2008 #3
    i tried using the conjugate way. The equation formed is extremely long @_@.
    Here is it:

    (x^2 - y^2 + 2x + 14yi + 48i) / (x^2 - y^2 - 36)

    There is still i in it @_@.
     
  5. Jun 12, 2008 #4

    Defennder

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    You could just substitute the z into the expression for w, multiply the denominator with its complex conjugate. You are given that w is imaginary, so what does that say about that expression you have?

    Anyway, I didn't manage to get that answer you gave. Did you transcribe the question correctly?
     
  6. Jun 12, 2008 #5

    tiny-tim

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    crays, there's supposed t be an i in it! :rolleyes:

    w has to be purely imaginary … that's zero plus something-times-i.

    If your equation were correct, the solution would be x² - y² + 2x = 0.

    You have a minus wrong, and the 48 seems to be in the wrong place …

    show us your working! :smile:
     
  7. Jun 12, 2008 #6
    w = imaginary means that the whole w equation = imaginary ? I don't really understand how does that helps tho. Erm yes, i'll repeat my question here again

    Given that z = x + yi and w = (z+8i)/(z-6) , z =/= 6 , show that x^2 + y^2 + 2x - 48 = 0
     
  8. Jun 12, 2008 #7

    HallsofIvy

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    You don't want to get rid of the w, you want to use the fact that it is "totally imaginary"-i.e., its real part is 0. If you put z= x+ yi into w, what is the real part of 0?
     
  9. Jun 12, 2008 #8
    Okay, my working :
    (z + 8i) / (z - 6) x (z + 6) / (z + 6)
    = (z² - 6z + 8zi + 48i) / (z² - 36)
    = [(x+yi)² - 6(x+yi) + 8(x+yi)i + 48i] / (x+yi)² - 36)
    = (x² - y² - 6x + 6yi + 8x + 8yi + 48i) / (x² - y² - 36)
    = (x² - y² + 2x + 14yi + 48i) / (x² - y² - 36)
     
  10. Jun 12, 2008 #9
    My teacher did not explain it properly. What does TOTALLY imaginary means? Means the whole equation is imaginary ? means there is no real number there? I'm confused @_@
     
  11. Jun 12, 2008 #10

    D H

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    Neither z+6 nor z-6 is not the complex conjugate of z-6. Your denominator is still complex.
     
  12. Jun 12, 2008 #11

    tiny-tim

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    crays, the conjugate of z - 6 is not z +6, is it? :frown:

    Hint: put z - 6 = x - 6 + yi. :smile:
     
  13. Jun 12, 2008 #12

    D H

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    An expression is totally imaginary or pure imaginary if the real part is identically zero.
     
  14. Jun 12, 2008 #13
    So sorry my maths foundation and expansion is still not good enough. Here is my new one, (hopefully without error)

    (x² - y² + 2x + 2yi - 48i) / (x² - y² - 36)

    D H @ I don't really get it, so 0 is imaginary ?
     
  15. Jun 12, 2008 #14

    D H

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    Show your work, please.

    The same question (is zero real or imaginary) can be applied to the real numbers: is zero positive or negative? Zero can be viewed as both positive and negative, or as neither positive nor negative. The same applies to complex numbers. Zero is a special case.
     
  16. Jun 12, 2008 #15

    tiny-tim

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    (x² - y² - 36) is wrong.

    Hint: what is (x - 6 - yi)(x - 6 + yi) ? :smile:
     
  17. Jun 12, 2008 #16
    Sorry, here is my working.

    [ (Z + 8i) / ( z - 6 ) ] x [ (z - 6) / (z - 6) ]
    = (z² - 6z + 8zi - 48i) / (z² - 36)
    = (x² - y² - 6x - 6yi + 8x + 8yi - 48i) / ( x² - y² - 36 )
    = (x² - y² + 2x + 2yi - 48i) / (x² - y² - 36)
     
  18. Jun 12, 2008 #17

    tiny-tim

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    crays, both D H and I have told you that z+6 is not the complex conjugate of z-6.

    Follow the hint! :smile:
     
  19. Jun 12, 2008 #18
    i am, just that i posted before you posted lol. Thanks.
    here it is, following ur hints, i've got.

    (x² - y² + 2x + 2yi - 48i) / (x² - 12x + 2xyi + 36 - 12yi - y²)
     
  20. Jun 12, 2008 #19

    tiny-tim

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    Nooo. Let's concentrate on the denominator …

    it should be purely real, shouldn't it?

    The x² - 12x + 36 is correct, but the y part shouldn't have any i, should it? :smile:
     
  21. Jun 12, 2008 #20
    finally!!! thanks.

    I've got it cause in z there is no imaginary number so i must expand it first right?
    with that i've found (denominator)
    x² - 12x + 36 + y²
     
    Last edited: Jun 12, 2008
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