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Math Help - Complex Number problem

  1. #1
    Newbie Bluedot's Avatar
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    Question Complex Number problem

    I need a solution to this:

    (problem attached by image)

    P.S. 'z' is in the form z = x + iy
    Attached Thumbnails Attached Thumbnails Complex Number problem-problem.jpg  
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  2. #2
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    Dear Bluedot,

    I have attached the solution to this question. If you find the attachment unreadable or if you have any questions regarding this solution please reply to me.

    Hope this helps.

    This solution is wrong. Please don't refer it.
    Attached Thumbnails Attached Thumbnails Complex Number problem-dsc02526.jpg  
    Last edited by Sudharaka; January 4th 2010 at 07:52 PM.
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  3. #3
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    Quote Originally Posted by Sudharaka View Post
    Dear Bluedot,

    I have attached the solution to this question. If you find the attachment unreadable or if you have any questions regarding this solution please reply to me.

    Hope this helps.
    There are a few mistakes in your solution.

    First, it is not necessarily correct that |z| \geq |z-1|: take, for example, z = -1 \Rightarrow |z| = 1, ~ \text{but } |z-1| = |-2| = 2 > 1 = |z|.

    Second, the fact that |arg z| \geq 0 does not justify keeping the inequality sign. For example, if we have a=2, b=3, c=\frac{1}{2} then a \leq b, but a > bc = \frac{3}{2}.

    Third, I believe he was looking for a specific solution (finding for which z this inequality holds), rather than a proof that this statement is true.
    Last edited by Defunkt; January 4th 2010 at 03:17 PM.
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  4. #4
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    Dear Defunkt,

    Yes you are correct. Thank you for showing the mistakes of my solution. As you said he may want a solution for which the inequality is valid. Since I cannot delete the thread I have marked it wrong.
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  5. #5
    Newbie Bluedot's Avatar
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    Third, I believe he was looking for a specific solution (finding for which z this inequality holds), rather than a proof that this statement is true.
    I am looking for a proof to the statement for any value of z (not any specific solution).
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  6. #6
    Senior Member Shanks's Avatar
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    If \left|z\right|\geq 1, then it is equivalent to prove that
    \left|\frac{1}{z}\right|+\left|1-\frac{1}{z}\right|\leq 1+\left|arg z\right|
    by letting w=\frac{1}{z} , then \left|w\right|\leq 1, the above expression can be rewrited as
    \left|w\right|+\left|1-w\right|\leq 1+\left|arg w\right|
    This is obviously true from the view of geometry (length) and the triangle inequality.
    similarly for the case \left|z\right|\leq 1.
    Last edited by Shanks; January 6th 2010 at 04:40 AM.
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  7. #7
    Newbie Bluedot's Avatar
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    Quote Originally Posted by Shanks View Post
    similarly for the case \left|z\right|\leq 1.
    If z = x + iy, where x and y are real, then |z| can never be less than 1, since z = sq.root(x^2+y^2)

    (sorry, don't know how to use latex)
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  8. #8
    Senior Member Shanks's Avatar
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    your claim is obviously false.
    two simple counterexample: z=0, \text{ or }z=\frac{1}{2}i.

    Can you show your effort on this prolem before asking help from other guys?
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