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Math Help - Proving for complex numbers

  1. #1
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    Proving for complex numbers

    How would you prove that z - 1/z = 2*j*sin(theta) ?
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  2. #2
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    Re: Proving for complex numbers

    Quote Originally Posted by yugimutoshung View Post
    How would you prove that z - 1/z = 2*j*sin(theta) ?

    If do you mean prove, then you must have more conditions on z.
    Like |z|=1.
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  3. #3
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    Re: Proving for complex numbers

    I think "z" refers to r*cis(theta)
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    Re: Proving for complex numbers

    Quote Originally Posted by yugimutoshung View Post
    I think "z" refers to r*cis(theta)
    If r\ne 1 then z-\frac{1}{z}\ne 2i\sin(\theta).

    So there is nothing to prove.
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    Re: Proving for complex numbers

    Quote Originally Posted by Plato View Post
    If r\ne 1 then z-\frac{1}{z}\ne 2i\sin(\theta).

    So there is nothing to prove.

    if z =r(cosθ+isinθ) then z^n=r^n[cos(nθ)+isin(nθ)] and z^(-n)=r^n[cos(-nθ)+isin(-nθ)]
    Therefore z^n+z^(-n)=2rcos(nθ) and z^n-z^(-n)=2risin(nθ)
    for n=1 and r=1 .........obvious

    check what you need.
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    Re: Proving for complex numbers

    Quote Originally Posted by MINOANMAN View Post
    if z =r(cosθ+isinθ) then z^n=r^n[cos(nθ)+isin(nθ)] and z^(-n)=r^n[cos(-nθ)+isin(-nθ)]
    Therefore z^n+z^(-n)=2rcos(nθ) and z^n-z^(-n)=2risin(nθ)
    for n=1 and r=1 .........obvious
    I don't understand if you meant comment for me or for the OP.

    As I said it is true if |z|=1 it is true. Here is the standard proof.

    Recall that if z=\text{c}i\text{s}(\theta) then \frac{1}{z}=\text{c}i\text{s}(-\theta).

    Now it is just a exercise to finish.
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