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Math Help - Natural Logarithm of a Complex Number

  1. #1
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    Natural Logarithm of a Complex Number

    I'm not sure how to go about solving this little question...

    By expressing z = i*r*e^(i*theta) in terms of 'z'*e^(i*theta) ('z' is the modulus of z), find the natural logarithm of z.

    I have thought of a couple of different approaches but the answers end up looking quite inelegant, so I think they may be wrong... and there's no answer in the back of the book...

    Thank you very much in advance!

    Jessica.
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  2. #2
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    I may be missreading your question.
    But the usual way of doing this is: \log (z) = \ln \left( {\left| z \right|} \right) + i\arg (z).

    Is that close to what ou mean?
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  3. #3
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    Yes, I know that's how the general result looks, but I was wondering what i*r*exp(i*theta) looks like in this form. I have a result, and it looks like this:

    log z = ln r + i(theta + pi/4 + 2pi*n)

    However I don't think this is right.
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  4. #4
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    Because i = e^{i\left( {\frac{\pi }{2}} \right)} then ire^\Theta   = re^{i\left( {\Theta  + \frac{\pi }{2}} \right)} .
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  5. #5
    Moo
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    Hello,
    Quote Originally Posted by j_clough View Post
    Yes, I know that's how the general result looks, but I was wondering what i*r*exp(i*theta) looks like in this form. I have a result, and it looks like this:

    log z = ln r + i(theta + pi/4 + 2pi*n)

    However I don't think this is right.
    The incorrect part is pi/4.

    Note that i=0+1 \times i=\cos \tfrac \pi 2+i \sin \tfrac \pi 2=e^{i \frac \pi 2}
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  6. #6
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    Ahh, I didn't see that! Thank you very much, Moo.
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