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Math Help - Seems easy but..

  1. #1
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    Seems easy but..

    I have this question I am suppose to solve and I think i misunderstood something. It is in a University algebra book and the questions is:
    Show that if abs:z =1 then, abs:  \left[\frac{2z-1}{z-2}\right]=1
    I dont see what is so difficult here?? why, why put this into a University book??
    If abs:z=1 then, abs:2z-1=1 and in the nominator abs:z-2 = -1 and absolute value of that is 1. So it is \frac{1}{1}?
    Have I missunderstood something?
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  2. #2
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    If |z| = 1, then z = -1 or z = 1. You must show that \left|\frac{2z - 1}{z - 2}\right| = 1 for both values of z.
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  3. #3
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    Quote Originally Posted by Henryt999 View Post
    I have this question I am suppose to solve and I think i misunderstood something. It is in a University algebra book and the questions is:
    Show that if abs:z =1 then, abs:  \left[\frac{2z-1}{z-2}\right]=1
    I dont see what is so difficult here?? why, why put this into a University book??
    If abs:z=1 then, abs:2z-1=1 and in the nominator abs:z-2 = -1 and absolute value of that is 1. So it is \frac{1}{1}?
    Have I missunderstood something?
    This is surely a question about complex numbers. If |z| = 1 then z can be any point on the unit circle in the complex plane. The question is asking you to show that in that case |z-2| = 2|z-\tfrac12|.
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