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Math Help - Show that question regarding complex numbers

  1. #1
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    Show that question regarding complex numbers

    The question:

    Show that Re(\frac{1 - z}{1 + z}) = 0 for any complex number z with |z| = 1.

    My attempt:
    I tried finding the the result of the division, hoping to get a value of 0 as the real part, to confirm the problem. This is what I got:

    Let z = a+ib
    \frac{1 - (a + ib)}{1 + (a + ib)} = \frac{(1 - a) - ib)}{(a + 1) + ib}
    \frac{(1 - a) - ib)}{(a + 1) + ib} . \frac{(a + 1) - ib)}{(a + 1) - ib} = \frac{[(1 - a) - ib)][(a + 1) - ib]}{(a + 1)^2 + b^2}
    \frac{(1 - a^2) - 2ib -b^2)}{(a + 1)^2 + b^2} = \frac{(1 - a) - 2ib -b^2}{(a + 1) + b^2}
    \frac{(1 - a) - b^2 - 2ib}{(a + 1) + b^2}
    Therefore the real part is:
    \frac{(1 - a) - b^2}{(a+1) + b^2}

    Now I'm not sure where to go from here. I think what I've done so far is correct. Any help would be great, thanks!
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  2. #2
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    Quote Originally Posted by Glitch View Post
    The question:

    Show that Re(\frac{1 - z}{1 + z}) = 0 for any complex number z with |z| = 1.

    My attempt:
    I tried finding the the result of the division, hoping to get a value of 0 as the real part, to confirm the problem. This is what I got:

    Let z = a+ib
    \frac{1 - (a + ib)}{1 + (a + ib)} = \frac{(1 - a) - ib)}{(a + 1) + ib}
    \frac{(1 - a) - ib)}{(a + 1) + ib} . \frac{(a + 1) - ib)}{(a + 1) - ib} = \frac{[(1 - a) - ib)][(a + 1) - ib]}{(a + 1)^2 + b^2}
    \frac{(1 - a^2) - 2ib -b^2)}{(a + 1)^2 + b^2}

    Why are you dropping terms here?

    = \frac{(1 - a) - 2ib -b^2}{(a + 1) + b^2}
    \frac{(1 - a) - b^2 - 2ib}{(a + 1) + b^2}
    Therefore the real part is:
    \frac{(1 - a) - b^2}{(a+1) + b^2}

    Now I'm not sure where to go from here. I think what I've done so far is correct. Any help would be great, thanks!
    Note that since |z|=1, then

    a^2+b^2=1

    (Pythagoras' Theorem)
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  3. #3
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    I think you have a mistake either in your calculation or your writing. the final expression is:

    [LaTeX ERROR: Convert failed]

    Now, since the magnitude of z is 1, you have [LaTeX ERROR: Convert failed] . Thus the numerator simplifies to 0 and hence the statement is proven!

    Edit: Didn't see you post Archie Meade.
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  4. #4
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    Whoops, I cancelled the terms by accident. Now it makes sense! Thanks.
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  5. #5
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    Quote Originally Posted by Glitch View Post
    The question:

    Show that Re(\frac{1 - z}{1 + z}) = 0 for any complex number z with |z| = 1.

    My attempt:
    I tried finding the the result of the division, hoping to get a value of 0 as the real part, to confirm the problem. This is what I got:

    Let z = a+ib
    \frac{1 - (a + ib)}{1 + (a + ib)} = \frac{(1 - a) - ib)}{(a + 1) + ib}
    [tex]\frac{(1 - a) - ib)}{(a + 1) + ib} . \frac{(a + 1) - ib)}{(a + 1) - ib} = \frac{[(1 - a) - ib)][(a + 1) - ib]}{(a + 1)^2 + b^2}[/MATH
    \frac{(1 - a^2) - 2ib -b^2)}{(a + 1)^2 + b^2} = \frac{(1 - a) - 2ib -b^2}{(a + 1) + b^2}]
    Looks like an error here, perhaps a typo. " 1- a^2" in the denominator of the first fraction has become " 1- a". Also " (a+ 1)^2" in the denominator has become " (a+ 1)" but since a fraction is 0 if and only if the numerator is 0, the denominator is irrelevant.

    \frac{(1 - a) - b^2 - 2ib}{(a + 1) + b^2}
    Therefore the real part is:
    \frac{(1 - a) - b^2}{(a+1) + b^2}

    Now I'm not sure where to go from here. I think what I've done so far is correct. Any help would be great, thanks!
    The numerator shoud be 1- a^2- b^2- 2ib- and a^2+ b^2= 1.
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  6. #6
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    It wasn't a typo, I foolishly tried to cancel the terms. Ahh well, I'll learn with practise!
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  7. #7
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     \displaystyle\frac{{1 - z}}{{1 + z}} = \frac{{\left( {1 - z} \right)\left( {1 + \overline z } \right)}}{{\left| {1 + z} \right|^2 }} = \frac{{1 + \overline z  - z - \left| z \right|^2 }}{{\left| {1 + z} \right|^2 }} = \frac{{\text{Im} (z)i}}{{\left| {1 + z} \right|^2 }}
    It is clear that the real part is zero.
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