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Log of i

  1. Apr 24, 2007 #1
    Can you get logarithms of imaginary numbers?

    The reason I was wondering is because integrating 1/(1+x^2) WRT x.
    I know the answer is arctan(x), but how about breaking it into partial fractions by doing 1/(1-ix)(1+ix)?
     
  2. jcsd
  3. Apr 24, 2007 #2

    KTC

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    Yes, though it's a mutivalued function so it's not as simple as with natural number.
     
  4. Apr 24, 2007 #3

    quasar987

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    Let's try to find log(i), given that log is the inverse function of the exponential defined on the whole complex plane.

    So, log(i), whatever it is, should be a number such that exp(log(i))=i. Let's say log(i)=z=x+iy... then z is such that exp(z)=exp(x+iy)=exp(x)exp(iy)=exp(x)[cos(y)+isin(y)]=i. Well, x=0, y=pi/2 does the trick. But so does x=0, y=pi/2+2npi for any integer n. So log(i) is actually a multivalued function.
     
  5. Apr 25, 2007 #4

    Alkatran

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    Given a complex number x = a + bi we want to find its logarithm
    [tex]c + di = ln(a + bi)[/tex]

    Take the exponential of both sides
    [tex]e^{c + di} = e^{ln(a + bi)}[/tex]
    [tex]e^c e^{di} = a + bi[/tex]
    [tex]e^c (cos(\frac{d}{\pi}) + i sin(\frac{d}{\pi})) = a + bi[/tex]

    Separating the imaginary and real terms:
    [tex]e^c cos(\frac{d}{\pi}) = a[/tex]
    [tex]e^c sin(\frac{d}{\pi}) = b[/tex]

    To get d, divide one by the other:
    [tex]\frac{e^c}{e^c} \frac{sin(\frac{d}{\pi})}{cos(\frac{d}{\pi})} = \frac{b}{a}[/tex]
    [tex]tan(\frac{d}{\pi})} = \frac{b}{a}[/tex]
    [tex]d = tan^{-1}(\frac{b}{a}) \pi[/tex]

    To get c, use the absolute values and Pythagoras:
    [tex]e^c e^{di} = a + bi[/tex]
    [tex]|e^c e^{di}|^2 = |a + bi|^2[/tex]
    [tex]|e^c|^2 = a^2 + b^2[/tex]
    [tex]e^c = \sqrt{a^2 + b^2}[/tex]
    [tex]c = ln(\sqrt{a^2 + b^2})[/tex]

    Note that you can add/subtract 2pi to d to get more solutions, which is a typical side effect of using arctan.
     
    Last edited: Apr 25, 2007
  6. Jul 29, 2011 #5
    I haven't read everything you wrote jet...

    but I had an Problem too:

    i^0 = 1
    i^1 = i
    i^2 = -1
    i^3 = -i
    i^4 = 1
    ...

    It's going on like that,
    but if you make de lg of the rationaly numbers...

    lg(i^0)=lg(1) ____ lg(i^2)=lg(-1) ____ lg(i^4)=lg(1) ...

    then there will be...

    0*lg(i)=lg(1) ____ 2*lg(i)=lg(-1) _____ 4*lg(i)=lg(1) ...

    ____ 0=lg(1) ______ lg(i)=lg(-1)/2 _____ lg(i)=lg(1)/4 ...

    NOW IS THAT POSSIBLE???

    lg(1)=0 ... That's True...
    BUT:
    lg(-1)=????

    My first Problem...
    Just forget it for the moment...

    SO: lg(i)=0/4 => lg(i)=0 => lg(i)=lg(1) => i=1 !!!!!!!!

    Now where was I wrong...???
    or is it just that you can't work that way...


    as an other example...

    i^2= -1
    i = -/-1]


    i*i = -/ -1 ]*-/ -1 ]

    So...

    i*i = -/ -1*-1 ]
    i*i = -/ 1 ]
    i*i = 1 ???????

    Both time's i=1...
     
  7. Jul 30, 2011 #6

    The fallacy is that √(xy)=√x√y is generally valid only if atleast one of the two numbers x or y is positive. (From Wiki)
     
  8. Jul 30, 2011 #7
    [tex]
    \log{(i)} = i \, \left(\frac{\pi}{2} + 2 \, k \, \pi\right), \; k \in \mathbb{Z}
    [/tex]
     
  9. Jul 31, 2011 #8
    You don't actually need to find ln(i) in order to perform the integration using partial fractions, but I think you will need the following formula:

    [tex]\tan^{-1}(z) = \frac{1}{2i} \ln \frac{1-iz}{1+iz}[/tex]

    You might want to see if you can derive it (it's not hard).
     
  10. Apr 27, 2013 #9
    log (i) = ?

    log (i) = x

    convert imaginary to exponential form...

    thus;

    (r)(exp^i) => i = (1)exp^(∏/2)i

    then substitute to equation...

    log (1exp^(∏/2)i)

    basic logarithmic law...

    log (ab) = log a + log b

    therefore...

    log (1exp^(∏/2)i) = log 1 + log exp^((∏/2)i)

    log 1 = 0

    log exp^((∏/2)i) = i log exp^(∏/2) = 0.6822 i

    therefore...

    log (i) = 0.6822i
     
  11. Apr 27, 2013 #10
    another way to solve log (i)

    log (i) = ?

    let log (i) = x

    therefore...

    10^x = i

    take the natural logarithm...

    ln 10^x = ln i

    ln 10^x = x ln 10

    ln (i) = ln (1 exp^((∏/2)i) = ln 1 + ln exp^(∏/2)i = (∏/2)i

    x ln 10 = (∏/2)i

    x = (∏/2)i / ln 10

    x = 0.6822 i

    same as the answer of my first post...
    different solution but one answer...
     
  12. Apr 27, 2013 #11

    pwsnafu

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    If you want a unique answer you need to pick a branch cut first.
     
  13. Apr 28, 2013 #12
    There are logarithms of [itex]i[/itex], but in terms of the calculus, your problem asks you to differentiate with respect to a [itex]x \in \mathbb{R}[/itex], right?

    While the complex numbers contains all the real numbers, you may need to justify why you can get a solution of a real variable using a method with complex numbers.
     
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