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Math Help - Limit of a complex variable

  1. #1
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    Limit of a complex variable

    lim z to i (z^2+i)/(z^4-1)=?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Take into account that is z^{4}-1= (z^{2}+1) (z^{2}-1)= (z-i) (z+i) (z-1) (z+1)...

    Kind regards

    \chi \sigma
    Last edited by chisigma; March 29th 2011 at 05:57 AM. Reason: see tonio's post...
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  3. #3
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    Yes, I did the same and I've found:

    lim z->i (z^2+i)/(z^4-1) = lim z->i (1/(z^2-i)) = 1/(i^2-i) = 1/(-1-i) = (-1/((1+i)) * (1-i/(1-i)) = (-1+i)/(1-i^2) = 1+i/(1+1) = (1+i)/2

    but in the solution manual it says the answer is: -1/2
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  4. #4
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    Quote Originally Posted by chisigma View Post
    Take into account that is z^{4}-1= (z^{2}-i) (z^{2}+i)...


    I'm afraid it is not: (z^2+i)(z^2-i)=z^4-i^2=z^4+1 ...A little blunder, but

    what is true is z^4-1=(z^2-1)(z^2+1)=(z-1)(z+1)(z-i)(z+i)

    Tonio




    Kind regards

    \chi \sigma
    .
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  5. #5
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    Quote Originally Posted by essedra View Post
    Yes, I did the same and I've found:

    lim z->i (z^2+i)/(z^4-1) = lim z->i (1/(z^2-i)) = 1/(i^2-i) = 1/(-1-i) = (-1/((1+i)) * (1-i/(1-i)) = (-1+i)/(1-i^2) = 1+i/(1+1) = (1+i)/2

    but in the solution manual it says the answer is: -1/2


    The solution manual is wrong unless you miswrote the numerator and it actually is z^2+1. As you

    wrote it, the limit doesn't exist.

    Tonio
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  6. #6
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    I've also checked out the cramster.com solution, in the nominator they've written z^2+1, I and found -1/2. So, you're right that there's an error whether in the question or the solution manual. You helped me to correct my misunderstanding, anyway. Many thanks for your help.
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