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Math Help - Proving a complex no

  1. #1
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    Proving a complex no

    How do I prove that

    |z/w| = |z|/|w| ?


    I had let z = a+bi and w = c+di so a+bi/c+di
    = a/c + (b/d)i
    So modulus is sqrt( (a/c)^2 + (b/d)^2)
    And then..I have a feeling I've gone really wrong here.

    Thank you very much for your help and time!
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  2. #2
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    Re: Proving a complex no

    Uhm... O.o

    If z=a+ib and w=c+id, then \frac{a+ib}{c+id} is not necessarily equal with \frac{a}{c}+\frac{b}{d}i.


    \frac{z}{w}=\frac{a+ib}{c+id}=\frac{(a+ib)(c-id)}{c^2+d^2}=\frac{ac+bd+(bc-ad)i}{c^2+d^2}

    \Rightarrow \left | \frac{z}{w} \right |=\sqrt{\left ( \frac{ac+bd}{c^2+d^2}^{} \right )^2+\left ( \frac{bc-ad}{c^2+d^2} \right )^2}=\sqrt{\frac{a^2c^2+b^2d^2+b^2c^2+a^2d^2}{(c^2  +d^2)^{2}}}

    =\sqrt{\frac{a^2(c^2+d^2)+b^2(c^2+d^2)}{(c^2+d^2)^  2}}=\sqrt{\frac{(a^2+b^2)(c^2+d^2)}{(c^2+d^2)^2}}=  \sqrt{\frac{a^2+b^2}{c^2+d^2}}=\frac{\sqrt{a^2+b^2  }}{\sqrt{c^2+d^2}}=\frac{|z|}{|w|}
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  3. #3
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    Re: Proving a complex no

    Quote Originally Posted by Tutu View Post
    How do I prove that
    |z/w| = |z|/|w| ?
    First you need to know (or prove) that:
    |\overline{z}|=|z| and |zw|=|z||w|.

    Have you done those two?
    Thanks from topsquark
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  4. #4
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    Re: Proving a complex no

    Thank you so so much!
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