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Thread: Complex Proof

  1. #1
    Member jacs's Avatar
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    Complex Proof

    $z\in C$ such that $\frac{z}{z-i}$ is real. Show that z is imaginary.

    just at the first stage of intro to complex numbers and beginning to struggle
    so far i have this but don't know how to proceed any further


    $\frac{z}{z-i}=\frac{z-i+i}{z-i}$
    $=\frac{z-i}{z-i}+\frac{i}{z-i}$
    $=1+\frac{i}{z-i}$
    $=1+\frac{i\times i}{i(z-i)}$
    $=1-\frac{1}{iz+1}$
    hence $\frac{1}{iz+1}$ is real

    multiply by conjugate
    $=\frac{1}{iz+1}\times\frac{1-i\overline{z}}{1-i\overline{z}}$
    $=\frac{1-i\overline{z}}{(iz+1)(-i\overline{z}+1)}$ which is real
    hence $i\overline{z}$ must also be real

    not quite sure what to do to proceed now.

    any help is enormously appreciated
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  2. #2
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    Re: Complex Proof

    Quote Originally Posted by jacs View Post
    $z\in C$ such that $\frac{z}{z-i}$ is real. Show that z is imaginary.
    just at the first stage of intro to complex numbers and beginning to struggle
    so far i have this but don't know how to proceed any further.
    We can show that \frac{z}{{z - i}} = \frac{{z(\overline z  + i)}}{{{{\left| {z - i} \right|}^2}}} = \frac{{z\overline z  + iz}}{{{{\left| {z - i} \right|}^2}}}.
    Because \frac{{z\overline z  + iz}}{{{{\left| {z - i} \right|}^2}}} is real conclude that $z$ must be imaginary.
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  3. #3
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    Re: Complex Proof

    let $z = a+bi$

    $\dfrac{z}{z-i} = \dfrac{a+bi}{a+(b-1)i} \cdot \dfrac{a-(b-1)i}{a-(b-1)i} = \dfrac{a^2+abi -a(b-1)i + b(b-1)}{a^2+(b-1)^2} \implies ab = a(b-1) \implies a = 0 \implies z = bi \, , \, b \ne 0$
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  4. #4
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    Re: Complex Proof

    I would simply have written \frac{z}{z- i}= a where a is real. Then z= a(z- i)= az- ai, (a- 1)z= ai so z= \frac{ai}{a- 1}= \frac{a}{a- 1}i. Since a is real, so is \frac{a}{a- 1}.
    Last edited by HallsofIvy; Jan 4th 2017 at 05:41 PM.
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