Results 1 to 8 of 8

Math Help - Natural log of a negative number

  1. #1
    Newbie
    Joined
    Sep 2007
    Posts
    17

    Natural log of a negative number

    I don't understand how to grasp such a concept. Obviously the answer is in a+bi form, but I have no idea how to get there.

    ln(-.3675)
    ln(-13.4)

    Here are a few problems from my book, can anyone guide me through?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Failbait View Post
    I don't understand how to grasp such a concept. Obviously the answer is in a+bi form, but I have no idea how to get there.

    ln(-.3675)
    ln(-13.4)

    Here are a few problems from my book, can anyone guide me through?
    If z\not = 0 then \log z = \log |z| + i\arg (z)

    Can you do that?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2007
    Posts
    17
    ln(-.3675) = -1.001+\pi i
    ln(-13.4) = 2.595+\pi i

    Do those answers look about right?

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Failbait View Post
    ln(-.3675) = -1.001+\pi i
    ln(-13.4) = 2.595+\pi i

    Do those answers look about right?

    Thanks!
    Hmmmm, let's see now ..... does |-0.3675| = -1.001? Does |-13.4| = 2.595?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    e^{i\theta} = cos \theta + i sin \theta
    e^{i\pi} = cos \pi + i sin \pi
    e^{i\pi} = -1
    ln(e^{i\pi}) = ln -1
    i\pi = ln -1

    For x \in  R^+
    ln -x = ln -1 + ln x = ln x + i\pi

    This is the same thing as ThePerfectHacker's formula. Remember that mine only works for real numbers.

    ln(-.3675) = -1.001+\pi i
    ln(-13.4) = 2.595+\pi i

    Yes, they're correct.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by wingless View Post
    ln(e^{i\pi}) = ln -1
    i\pi = ln -1
    You are not allowed to do that. The rule \log e^z = z for z\in \mathbb{C}^{\text{x}} does not work in general.
    However, e^{\log z} = z does work in general.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    Quote Originally Posted by ThePerfectHacker View Post
    You are not allowed to do that. The rule \log e^z = z for z\in \mathbb{C}^{\text{x}} does not work in general.
    Wow, I didn't know that. Sorry for that mistake, thanks for correcting me. Ok, the formula is still true since ln -1 = i \pi.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by wingless View Post
    Ok, the formula is still true since ln -1 = i \pi.
    Yes! Sometimes the formula will work sometimes it does not.
    Another thing that breaks down is \log (z_1z_2) = \log z_1 + \log z_2.
    Take for example z_1=z_2=-1.
    Complex logarithm.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: June 27th 2011, 08:25 PM
  2. Negative number in the denominator
    Posted in the Algebra Forum
    Replies: 4
    Last Post: September 18th 2010, 09:24 AM
  3. Replies: 6
    Last Post: February 6th 2010, 08:09 PM
  4. Replies: 10
    Last Post: April 8th 2009, 11:56 AM
  5. The natural number p is prime
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: February 11th 2009, 08:16 AM

Search Tags


/mathhelpforum @mathhelpforum