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Thread: complex logarythm function

  1. #1
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    complex logarythm function

    Hey
    let z1,z2...,zn be complex numbers in the right plane such that their product is also in the right plane.prove that:

    Log(z1*z2*...*zn)=Log(z1)+Log(z2)+...Log(zn)

    where Log(z) denotes the principal branch of the logarythm function.
    Thank's in advance
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  2. #2
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    Re: complex logarythm function

    Recall that if $\displaystyle \begin{align*} z = r\,\mathrm{e}^{\mathrm{i}\,\theta } \end{align*}$ then $\displaystyle \begin{align*} \log{ \left( z \right) } = \log{ \left( r \right) } + \mathrm{i}\,\theta \end{align*}$, so

    $\displaystyle \begin{align*} \log{ \left( z_1 \right) } + \log{\left( z_2 \right) } + \log{ \left( z_3 \right) } + \dots + \log{ \left( z_n \right) } &= \log{ \left( r_1 \right) } + \mathrm{i}\,\theta_1 + \log{ \left( r_2 \right) } + \mathrm{i}\,\theta_2 + \log{ \left( r_3 \right) } + \mathrm{i}\,\theta_3 + \dots + \log{ \left( r_n \right) } + \mathrm{i}\,\theta_n \\ &= \log{ \left( r_ 1 \right) } + \log{ \left( r_2 \right) } + \log{ \left( r_3 \right) } + \dots + \log{ \left( r_n \right) } + \mathrm{i}\,\theta_1 + \mathrm{i}\,\theta_2 + \mathrm{i}\,\theta_3 + \dots + \mathrm{i}\,\theta_n \\ &= \log{ \left( r_1 \cdot r_2 \cdot r_3 \cdot \dots \cdot r_n \right) } + \mathrm{i}\,\left( \theta_1 + \theta_2 + \theta_3 + \dots + \theta_n \right) \\ &= \log{ \left( z_1 \cdot z_2 \cdot z_3 \cdot \dots \cdot z_n \right) } \end{align*}$
    Last edited by Prove It; Dec 19th 2017 at 05:04 PM.
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  3. #3
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    Re: complex logarythm function

    Log here is the principle bruch of the logarithm function so the sum of the angles imust lie between -pi and pi.I think that it should be given that z1*z2*...*zk lies in the right half plane for all k's between 1 and n,otherwise the statement is false.Then the statement follows by induction.Do you agree?
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