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Math Help - arg(\frac{1}{z})

  1. #1
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    arg(\frac{1}{z})

    arg(z)=\theta+2\pi k, \ k\in\mathbb{Z}

    arg\left(\frac{1}{z}\right)=arg\left(\frac{\overli  ne{z}}{z\overline{z}}\right)
    =arg\left(\frac{\overline{z}}{|z|^2}\right)=arg(\o  verline{z})-arg(|z|^2)
    =-\theta-(\theta+2\pi k)=-2\theta-2\pi k

    Is this correct?
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  2. #2
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    Quote Originally Posted by dwsmith View Post
    arg(z)=\theta+2\pi k, \ k\in\mathbb{Z}

    arg\left(\frac{1}{z}\right)=arg\left(\frac{\overli  ne{z}}{z\overline{z}}\right)
    =arg\left(\frac{\overline{z}}{|z|^2}\right)=arg(\o  verline{z})-arg(|z|^2)
    =-\theta-(\theta+2\pi k)=-2\theta-2\pi k

    Is this correct?
    Note that \displaystyle \frac{1}{z} = \frac{\text{cis} (0)}{r \text{cis} (\theta)} = \frac{1}{r} \text{cis} (0 - \theta).
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  3. #3
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    So the the argument of 1 over z is just going to be -\theta+2\pi k?
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  4. #4
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    Letís consider the principal argument, -\pi < \text{Arg}(z)\le \pi.
    The  \text{Arg}(\overline z) is simply its reflection in the real axis.

    You may want to reconsider?
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  5. #5
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    If we pick up from where Mr F stopped, we have the angle as 0-\theta=-\theta.

    Now, my question is then why isn't the arg\left(\frac{1}{z}\right)=-\theta+2\pi k \ \mbox{?}
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by dwsmith View Post
    arg(z)=\theta+2\pi k, \ k\in\mathbb{Z}

    arg\left(\frac{1}{z}\right)=arg\left(\frac{\overli  ne{z}}{z\overline{z}}\right)
    =arg\left(\frac{\overline{z}}{|z|^2}\right)=arg(\o  verline{z})-arg(|z|^2)
    =-\theta-(\theta+2\pi k)=-2\theta-2\pi k

    Is this correct?
    Given  c\in\mathbb{R} and  z\in\mathbb{C} then  \text{arg}(cz)=\text{arg}(z) .

    What does this say about  \text{arg}\left(\frac{\overline{z}}{|z|^2}\right) \; ?
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  7. #7
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    arg\left(\frac{\overline{z}}{|z|^2}\right)=arg(\ov  erline{z})=-\theta+2\pi k, yes?
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  8. #8
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by dwsmith View Post
    arg\left(\frac{\overline{z}}{|z|^2}\right)=arg(\ov  erline{z})=-\theta+2\pi k, yes?
    Yes, since z is reflected about the x-axis.
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  9. #9
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    I think that
    <br />
arg(|z|^2)=arg(real \; number)=0<br />

    <br />
arg(1/z)=arg(\frac{1}{r} \; e^{-if})=-f+2 \pi n<br />
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