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Thread: Complex function question

  1. #1
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    Complex function question

    I'm having problem understanding how to get to this solution.

    log(-1+0i)=ipi

    Thank you for your help!
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  2. #2
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    Re: Complex function question

    There's no solution to this expression. It's an identity.

    $\text{ by Euler's formula.}$

    $-1 +0i= \cos(\pi) + i \sin(\pi) = e^{i \pi} $

    $\text{taking logs on both sides}$

    $\log_e(-1+0i) = i \pi$
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  3. #3
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    Re: Complex function question

    Thank you!
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  4. #4
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    Re: Complex function question

    Quote Originally Posted by mathfn View Post
    I'm having problem understanding how to get to this solution.
    log(-1+0i)=ipi
    I would expect a student to use the properties of logarithm: $\log(z)=\log(|z|)+{\bf{i}}\arg(z)$
    $\log(|-1|)=\log(1)+\bf{i}\pi$.
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