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Math Help - complex limit

  1. #1
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    complex limit

    Show that
    lim_{|z|\rightarrow 0} \frac{ln(z)}{z^2+a^2} = 0
    where z is a complex number
    Thanks
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  2. #2
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    Re: complex limit

    Quote Originally Posted by hurz View Post
    Show that
    lim_{|z|\rightarrow 0} \frac{ln(z)}{z^2+a^2} = 0
    where z is a complex number
    Thanks

    Use L´Hospital's rule , as Ln z=\ln |z|+i\arg (z) \xrightarrow [|z|\to 0] .  -\infty \, , \,z^2+a^2\xrightarrow [|z|\to 0]{}{}   \infty

    Tonio
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  3. #3
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    Re: complex limit

    Quote Originally Posted by tonio View Post
    Use L´Hospital's rule , as Ln z=\ln |z|+i\arg (z) \xrightarrow [|z|\to 0] .  -\infty \, , \,z^2+a^2\xrightarrow [|z|\to 0]{}{}   \infty

    Tonio
    I'm sorry, i'm not getting why
    \,z^2+a^2\xrightarrow [|z|\to 0]{}{}   \infty.

    Why it isn't
    \,z^2+a^2\xrightarrow [|z|\to 0]{}{} a^2 ?
    In the cause above, i could not use L'Hospital's Rule.
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  4. #4
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    Re: complex limit

    If i want to show that
    lim_{|z|\rightarrow 0} |\frac{ln(z)}{z^2+a^2}| = 0
    does it get easier?
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  5. #5
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    Re: complex limit

    Quote Originally Posted by hurz View Post
    I'm sorry, i'm not getting why
    \,z^2+a^2\xrightarrow [|z|\to 0]{}{}   \infty.

    Why it isn't
    \,z^2+a^2\xrightarrow [|z|\to 0]{}{} a^2 ?
    In the cause above, i could not use L'Hospital's Rule.
    Regards
    Of course you´re right, my bad. Mistook the limit to infinity instead of zero in the denominator.

    Tonio
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