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Logs of negative numbers (any base)

  1. Feb 6, 2006 #1
    I know that with base e you use the natural log of complex numbers rule...

    ln(x+yi) = ln|x| + arcsin(y/x)i

    ...because the limit e somehow corresponds to radians about the unit circle

    thus, when you take ln(-1) you can do

    ln(-1 + 0i) = ln|-1| = arsin(0/-1)i
    ln(-1) = 0 + pi*i
    ln(-1) = pi*i

    My first question is this: What is the relationship between the limit e as the log's base and the unit circle with respect to complex answers?

    My second question: How do you find the log of a negative number when the base is not the limit e?

    aside... it just dawned upon me while typing this that i might be able to jsut devide the complex answer by the ln of the desired base? such that...

    logb(-1) = ln|-1|/ln(b) + arctan(0/-1)i/ln(b)

    logb(-1) = logb|-1| + [1/ln(b)][arctan(0/-1)]i

    is that the proper way to do this?
  2. jcsd
  3. Feb 6, 2006 #2
    Remember the base change rule that you usually learn when you learn logarithms, you could use that to change to base e which I think is what you did.


    ln|x + iy | is not equal to ln|x| + arctan(y/x)i

    From my complex analysis book the natural log of a complex number z ( z = x + iy) is defined as

    log z = Log|z| + i*arg z (where |z| = sqrt(x2 + y2) )
    and this is equal to Log|z| + i*Arg z + i*2*pi*k where k is an integer.
    Last edited: Feb 6, 2006
  4. Feb 6, 2006 #3


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    The multivalued complex logarithm function is simply the inverse of the complex exponential. That is, [itex]\ln z[/itex] is the set of numbers that satisfy the equation [itex]\exp(w) = z[/itex].

    One can write an expression for ln z, but as d_leet pointed out, yours wasn't quite right. The usual way to write it is [itex]\ln z = \ln |z| + i \arg z[/itex], where the ln on the R.H.S. is the real number natural logarithm.

    There is a relationship between arg and arctan, but it isn't quite as trivial as the expression you wrote.

    Why is there a connection? Because of the famous formula [itex]\exp(iz) = \cos z + i \sin z[/itex]! The modern proof of this fact comes from the Taylor series for each side.

    But are we supposed to think of it as a magic formula that yields a mysterious connection between the disparite ideas of exponentials and trigonometry? No, not really. Complex analysis is a highly geometric subject -- we're "supposed" to think of multiplication by [itex]\exp(iz)[/itex] as being nothing other than a rotation about the origin by z radians. (Or, at least that's one of the ways we're supposed to think of it)

    As for other bases...

    You may have wondered why I did not write [itex]e^w = z[/itex] as my first LaTeX equation, or use exponentiation anywhere else before now -- that's because in the complex world, exponentiation is not a function: it is a multi-valued operation. For example, in the complex world both +1 and -1 are both values of [itex]1^{1/2}[/itex]. Unlike the real numbers, there is no "canonical" way to pick which of many options is "correct". (which leads to the notion of selecting a branch cut)

    e is special, because there is a way to choose a particular value of [itex]e^z[/itex] in a reasonable way: the value of the function [itex]\exp(z)[/itex]. So, it is reasonable to ask about its inverse, and thus arrive at the complex logarithm.

    However, for an arbitrary base, I'm not really sure how much use (or even sense) a logarithm would make. I don't think I've seen any other logarithm used in complex analysis.
  5. Feb 6, 2006 #4
    I thank you both... i did make a few typos in my original post such as using arcsin instead of arctan...

    anyway, in my calculus class we get extra credit if we can prove things... such as logb(a) and loga(b) are mutliplicative inverses...

    my teacher wants me to show how to find the log of any number with any base...

    so, that's why i'm interested in the relationships...

    i have found that i can find the log of any number with any base by doing the following...

    logb(x+yi) = logb|x| + [1/ln(b)][arctan(y/x)]i

    although, from what you said... i take it this is not quite right even though it yields the correct answer?
  6. Feb 6, 2006 #5
    The problem is with the range of the arctangent function it will always return a result between negative pi over two and pi over 2, but the arguments of numbers in the complex plane ( not sure if this is making sense I can try and explain better if it would help) are not limited to being between these values. Also you can't just take the log of the real part of the complex number, this works only if you deal purely with real numbers but once you start dealing with numbers such as -2 + 3i your formula won't work anymore.
    Last edited: Feb 6, 2006
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