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Complex analysis showing solutions are inside or outside R

  1. Apr 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose w is not in the interval [-R,R] show that the equation z+[itex]\frac{R^{2}}{z}[/itex]=2w has one solution z with |z|<R and one solution z with |z|>R

    2. Relevant equations

    none

    3. The attempt at a solution

    the book mentions that the quadratic is left unchanged by the substitution of [itex]\frac{R^{2}}{z}[/itex] for z which I understand.

    It then goes on to say that because of this, the two roots z[itex]_{1}[/itex] and z[itex]_{2}[/itex]are related by z[itex]_{1}[/itex]z[itex]_{2}[/itex]=R[itex]^{2}[/itex] which is where I get lost..

    any help is appreciated. Thanks
     
  2. jcsd
  3. Apr 8, 2012 #2

    Dick

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    Write your equation as z^2-2wz+R^2=0. If there are two roots z1 and z2 then it factors into (z-z1)(z-z2)=0. Multiply that back out and equate it to the original equation.
     
  4. Apr 8, 2012 #3
    ahh ok.. can't believe I missed that. now the book says that because z1z2=R^2 the roots lie on opposite sides of the circle |z|=R. Why is this? i'm assuming they mean inside the circle and outside the circle when they say opposite sides.. And my guess would be that z1z2 = R^2 when z1=z2=R or if z1 is larger than R and z2 is smaller than R or vice versa. The latter meaning that z1 is inside the circle and z2 is outside. The first would mean that the roots lie on the circle|z|=R?
     
  5. Apr 8, 2012 #4

    Dick

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    |z1|*|z2|=R^2. So if |z1|<R, |z2|>R. That's the easy case. You are done. Now suppose |z1|=|z2|=R. To eliminate that possibility you'd better say something about w.
     
  6. Apr 9, 2012 #5
    so for |z1|=|z2|=R, w would have to be in [-R,R] but this possibility was eliminated in the statement of the problem. So then the only possibility is for |z1|<R and |z2|>R or vice versa.
     
  7. Apr 9, 2012 #6

    Dick

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    How did you show w would need to be in [-R,R]?
     
  8. Apr 9, 2012 #7
    because for |z1|=|z2|=R would mean that (z-w)[itex]^{2}[/itex]=z[itex]^{2}[/itex]-2zw+w[itex]^{2}[/itex]=z[itex]^{2}[/itex]-2zw+R[itex]^{2}[/itex] so w[itex]^{2}[/itex]=R[itex]^{2}[/itex] and w=R which is in [-R,R]
     
  9. Apr 9, 2012 #8

    Dick

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    Why should (z-w)^2=z^2-2zw+R^2??? Your conclusion that w^2=R^2 is false. If you think harder about it you should be able to figure out the w is related to the sum of the roots z1 and z2, just like R^2 is related to the product of the roots.
     
  10. Apr 9, 2012 #9
    z1z2 must = R^2 and -z1-z2 must = -2w so z1+z2=2w now since |z1|=|z2|=R we have that R+R=2R=2w meaning R=w and R is in [-R,R]
     
  11. Apr 9, 2012 #10

    Dick

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    Yes, z1*z2=R^2 and z1+z2=2w. And yes, assume |z1|=|z2|=R. The rest of that doesn't make sense. w doesn't have to equal R.
     
  12. Apr 9, 2012 #11
    I thought if |z1|=|z2|=R then 2w=|z1|+|z2|=R+R=2R
     
  13. Apr 9, 2012 #12

    Dick

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    That doesn't work. Pick z1=iR and z2=(-iR). That makes w=0.
     
  14. Apr 9, 2012 #13
    ok.. so I know that z1*z2=R^2 and z1+z2=2w. And |z1|=|z2|=R I am just unsure of how to conclude that w is in [-R,R]...

    thanks for your patience with this, its just not clicking.
     
  15. Apr 9, 2012 #14

    Dick

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    Can you convince me that z1 and z2 are complex conjugates of each other?
     
  16. Apr 9, 2012 #15
    could we say that since z1*z2=R^2 and R is a real number, that z1 and z2 have to be of the form (a+ib)(c-id) to eliminate the imaginary parts and since we are saying that |z1|=|z2|=R we know that a=c and b=d in the equation so z1 and z2 must be complex conjugates.
     
  17. Apr 9, 2012 #16

    Dick

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    That's not at all convincing. What must z1 and z2 look like in polar form?
     
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