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Math Help - Complex Logarithms

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    is up to his old tricks again! Jhevon's Avatar
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    Complex Logarithms

    I'm not sure if this is the right forum to place this in, I know most of you on the forum should be able to deal with this kind of stuff in your sleep, but it's an Advanced Topic to me.

    About 10 minutes ago, for some reason unknown to me, I developed an interest in Complex Logarithms (I don't know if that's what it's called). I am talking about taking the logs of negative numbers and complex numbers. I browsed a few sites, including wikipedia, but most of the stuff flew over my head, I was hoping someone here could dumb it down for me. I will expound all that I have gotten from my "research" and hopefully someone can fill in the gaps.


    Here's some background knowledge for the small percentage of readers that actually know less about this topic than I do.


    Background:

    There is a nifty little thing that we call Euler's Formula, which can be derived by using Taylor Series (among other things I suppose). The formula states that:

     \boxed { e^{i \theta} = \cos \theta + i \sin \theta }

    where i is the imaginary unit, specifically, i = \sqrt {-1}

    Using this formula, we can derive what is perhaps the most aesthetic identity in all of Math, this identity is known as Euler's Identity.

    Euler's Identity:
    \boxed { e^{i \pi} + 1 = 0 }

    Or as it is usually stated:

    \boxed { e^{i \pi} = -1 }

    We obtain this beautiful identity by simply plugging in \pi for \theta in Euler's Formula.


    Another thing the reader should be aware of is how to represent a complex number in polar coordinates. This is (relatively) easy if we do so by the use of an Argand Diagram.

    Recall that a Complex Number is a number made up of a real number component and a imaginary number component. The sum of these components form the complex number. In an Argand diagram, we set up a vertical and a horizontal axis much like the familiar Cartesian axis', but with different meaning. The horizontal axis represents the real number line, while the vertical axis represents the imaginary number line. Any point on the imaginary axis is purely imaginary (that is, the real component of any complex number represented in this way is zero), any point on the horizontal axis is purely real (the imaginary part of the complex number is zero), and any point that is not on either axis represents a complex number, that is, one of its coordinates will represent the real part of the complex number (the first number in the coordinate pair), and the other coordinate represents the imaginary part of the complex number. For the sake of familiarity, we will use x to represent a number on the real (horizontal) axis and use the variable y to represent a number on the imaginary (vertical) axis.

    We define r to be the length of the line that connects the origin to any point in the Argand-plane. We can find the value of r using Pythagoras' Theorem. And we can find the angle that r makes with the real axis using the tangent ratio. In doing so, we can define any complex number much like the way we define Cartesian Coordinates in polar coordinates.

    Recall: To go from Polar Coordinates (r, \theta) to Cartesian Coordinates (x,y) , we use:

    x = r \cos \theta and y = r \sin \theta

    where r = \sqrt {x^2 + y^2} and \theta = \arctan \left( \frac {y}{x} \right)


    So, for any complex number x + iy, we can write

    x + iy = r \cos \theta + i~r \sin \theta

    \Rightarrow x + iy = r ( \cos \theta + i \sin \theta)

    But we can rewrite what's in brackets using Euler's Formula. Thus we see that the said complex number can be expressed as follows:

     \boxed { x + iy = r \cdot e^{i \theta} }

    where r = \sqrt {x^2 + y^2} and \theta = arctan \left( \frac {y}{x} \right)

    Finally we are done with introductions, let's get to the meat of the matter. Here are the scraps of knowledge that I could pick up.



    Taking logarithms of Negative Numbers

    So far as I've seen, we make much use of Euler's formula, and polar representation of complex number when dealing with complex logarithms, and so it comes as no surprise that the natural logarithm, \ln, is what we work with here.

    Complex Logarithms use an imaginary component analogous to that of i used to denote the imaginary component in a complex number. This component is i \pi and it is defined to be \ln (-1). This is an ingenious contraption, since any negative number can be thought of as -1 times the positive opposite of the negative number. Thus we can use the product law of logarithms (and if needed, the change of base formula for logarithms) to isolate the \ln (-1) from any log of a negative number.

    Before we see a demonstration of the technique, some readers may be wondering why is it that we use i \pi to mean \ln (-1). It is no big deal, we simply apply some laws of logarithms coupled with Euler's Identity (see the "Background" section above).

    Recall the laws of logarithms (I will number the laws to refer to them in this thread, but the numbers should not be taken to mean anything other than that, that is, don't assume that the law I call "Law 1" is a fundamental law of logarithms and is of huge importance. The laws are in no particular order):

    Law 1: \log_a x^n = n \log_a x

    Law 2: \log_a a = 1

    Law 3: We define \log_e x to be \ln x. Therefore, \ln e = 1 (by Law 2).

    Law 4: \log_a b = \frac { \log_c b}{ \log_c a} .....change of base law

    Law 5: \log_a xy = \log_a x + \log_a y


    Now we can examine \ln (-1)

    \ln (-1) = \ln \left( e^{i \pi} \right) ........Applied Euler's Formula

    \Rightarrow \ln (-1) = i \pi \ln e .............Applied Law 1

    \Rightarrow \ln (-1) = i \pi \cdot 1 ...............Applied Law 3

    Thus we have:

    \boxed { \ln (-1) = i \pi }


    Now that we have all doubt dispersed regarding our "imaginary" component (I'm not sure if that's what it's called), let us see it in action.

    Example 1:

    Evaluate \ln (-10)


    \ln (-10) = \ln (-1 \cdot 10)

    = \ln (-1) + \ln (10) .........Law 5

    = i \pi + \ln (10)

    \approx 2.303 + i \pi


    What if we have a logarithm in some other base? We simply use the change of base formula, that is, Law 4

    Example 2:

    Evaluate \log_{10} (-100)


    \log_{10} (-100) = \frac { \ln (-100) }{ \ln 10} ........Law 4

    = \frac { \ln (-1 \cdot 100)}{ \ln 10}

    = \frac { \ln (-1) + \ln 100}{ \ln 10} .............Law 5

    = \frac {i \pi + \ln 100}{ \ln 10}

    = \frac {i \pi}{ \ln 10} + \frac { \ln 100}{ \ln 10}

    = i \frac { \pi}{ \ln 10} + \frac { \ln 10^2}{ \ln 10}

    = i \frac { \pi}{ \ln 10} + \frac {2 \ln 10}{ \ln 10} .........Applied Law 1 to the numerator

    = i \frac { \pi}{ \ln 10} + 2 ...........Finally.

    This seems to take forever, but that's just because I showed every step. Textbooks often get criticized for not showing all the steps, and students get frustrated when they do not understand how the book jumped from one step to another. I didn't want such criticisms to fall on me. However, we could still make it shorter. We did not have to jump into the change of base formula like that, we could have waited until we isolated the \log_{10} (-1)

    Rework Example 2:

    \log_{10} (-100) = \log_{10} (-1 \cdot 100)

    = \log_{10} (-1) + \log_{10} (100) ..............Law 5

    = \frac { \ln (-1)}{ \ln 10} + 2 ..............Applied Law 4 to the first log while evaluating the second

    = \frac {i \pi}{ \ln 10} + 2

    = 2 + \frac { \pi}{ \ln 10}~i


    So yeah, we can take the logarithm of negative numbers, they lied to you! Just like they lied to you when you were a child telling you you cannot subtract a bigger number from a smaller number, and then you get to the 6th grade and they do it all the time!

    Anyway, I forgive them. Let's summarize what we have learnt here.

    Let x > 0 be a real number.

    Then \boxed { \ln (-x) = \ln x + i \pi }

    and

    \boxed { \log_a (-x) = \frac { \ln x}{ \ln a} + \frac { \pi}{ \ln a}~i }

    Now that we know we can take the logarithm of a negative number, the next logical question to ask (I think) would be: "If we can take the logarithm of a negative number to get a complex number, does that mean we can take the logarithm of a complex number as well?" (That's the question I asked myself when I researched the above, I don't know if it is a logical train of thought or not--generally I'm not a logical person).

    The answer is yes. Follow me deeper into the dark.



    Taking logarithms of Complex Numbers

    Recall the we could express any complex number, x + i y by the following:

     \boxed { x + iy = r \cdot e^{i \theta} }

    where r = \sqrt {x^2 + y^2} and \theta = arctan \left( \frac {y}{x} \right)

    With all the background knowledge I gave, I don't think it is hard for the average reader to see where I'm going with this, so I'll just say it.

    We can take the logarithm of a complex number by making use of the following formula:

    \boxed { \log_a (x + iy) = \log_a \left( r \cdot e^{i \theta} \right) }

    or equivalently (by applying Law 5): \boxed { \log_a (x + iy) = \log_a r + \log_a e^{i \theta} }

    or to simplify even more (by applying Law 1): \boxed { \log_a (x + iy) = \log_a r + i \theta \log_a e }

    Not much explanation is needed here. Let's jump right into the examples.

    Let's start with something "simple"

    Example 3:

    Evaluate \ln (1 + i)


    First we rewrite 1 + i in polar coordinates, that is, in the form r \cdot e^{i \theta}. Using the formulas discussed earlier in this thread, we find

    r = \sqrt {1^2 + 1^2} = \sqrt {2}

    \theta = \arctan 1 = \frac { \pi}{4}

    Thus,

    \ln (1 + i) = \ln \left( \sqrt {2} \cdot e^{i \pi /4}\right)

    = \ln \sqrt {2} + \ln e^{ i \pi /4}

    = \frac { \ln 2}{2} + i \frac { \pi}{4} \ln e ..........Applied Law 1 to both logs (remember \sqrt {2} = 2^{ \frac {1}{2}})

    = \frac { \ln 2}{2} + \frac { \pi}{4} ~i



    Let's do another.

    Example 4:

    Evaluate \ln (-4 + 3i)

    First we find r \mbox { and } \theta

    r = \sqrt {(-4)^2 + 3^2} = \sqrt {25} = 5

    \theta = \arctan \left( - \frac {3}{4} \right) \approx -0.644

    Thus,

    \ln (-4 + 3i) = \ln \left( 5 \cdot e^{-0.644i} \right)

    = \ln 5 + \ln e^{ -0.644i} ............Applied Law 5

    = \ln 5 - 0.644i \ln e

    = \ln 5 - 0.644i

    That wasn't so bad, was it?


    One more example for ol' times sake. Let's do one where we need the change of base law.


    Example 5:

    Evaluate \log_3 (-2 - 8i)

    Find r \mbox { and } \theta

    r = \sqrt {(-2)^2 + (-8)^2} = \sqrt {68}

    \theta = \arctan \left( \frac {-8}{-2} \right) \approx 1.326

    Thus

    \log_3 (-2 - 8i) = \log_3 \left( \sqrt {68} \cdot e^{1.326i} \right)

    = \log_3 \sqrt{68} + 1.366i \log_3 e .........Applied Law 1

    \approx 1.92 + 1.243i .............I applied law 4 (change of base law) to evaluate the logs in base 3



    So to summarize what we have learned here.

    Let x + iy be any complex number. And let r = \sqrt {x^2 + y^2} and \theta = \arctan \left( \frac {y}{x} \right)

    Then \boxed { \ln (x + iy) = \ln r + i \theta }

    and

    \boxed { \log_a (x + iy) = \log_a r + ( \theta \log_a e)~i }


    Exercises

    For those readers who would like to see if they really "get it," here are some problems you can practice on. You can post your solutions in this thread for the other members to critique... if you want. I made up random numbers, so don't expect the answers to be clever or "nice." (I could make some "nice" problems if I wanted to, but I'm not in a thinking mood at the moment).

    Evaluate the following logarithms:

    \ln (i)

    \log_7 (-49)

    \log_17 0

    \log_a \left( - \frac {4}{5} \right)

    \log_5 ( 4 + i)

    \log_3 (-6 - 7i)

    \log_2 (-3 + 1)

    \log_8 (8 - 3i)



    Despite appearances, I did not mean for this thread to be a tutorial of any sort, I made this thread so that the genius' on this forum can give me feedback as to whether or not I actually understand the material. Here are just a few questions I have at the moment, more questions may arise if any discussion results.

    Questions:

    - Was everything I said correct? Would anyone like to add anything that can deepen my, as well as other readers', understanding?

    - Is there a way to take logarithms of negative numbers without working with natural logarithm?

    - Doing all the things I said seems cumbersome, is there anyway to make it easier? Like some mnemonic I should know or something?

    - Are there logs with negative or complex bases? If so, how would we evaluate them?

    - (A bit off-topic) How do we evaluate numbers with complex exponents? (I think I could have come up with more interesting problems if I knew that)
    Last edited by Jhevon; July 1st 2007 at 12:38 AM. Reason: Fixed typo. Thanks JakeD
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    Quote Originally Posted by Jhevon View Post
    Taking logarithms of Complex Numbers

    Recall the we could express any complex number, x + i y by the following:

     \boxed { x + iy = r \cdot e^{i \theta} }

    where r = \sqrt {x^2 + y^2} and \theta = arctan \left( \frac {y}{x} \right)

    With all the background knowledge I gave, I don't think it is hard for the average reader to see where I'm going with this, so I'll just say it.

    We can take the logarithm of a complex number by making use of the following formula:

    \boxed { \log_a (x + iy) = \log_a \left( r \cdot e^{i \theta} \right) }

    or equivalently (by applying Law 5): \boxed { \log_a (x + iy) = \log_a r + \log_a e^{i \theta} }

    or to simplify even more (by applying Law 1): \boxed { \log_a (x + iy) = \log_a r + i \pi \log_a e }

    [snipped]

    Questions:

    - Was everything I said correct? Would anyone like to add anything that can deepen my, as well as other readers', understanding?
    First, there is a typo:

    \boxed { \log_a (x + iy) = \log_a r + i \pi \log_a e }

    should be

    \boxed { \log_a (x + iy) = \log_a r + i \theta \log_a e }.


    Second and more importantly, the \theta is not uniquely defined in

     \boxed { x + iy = r \cdot e^{i \theta} }

    where r = \sqrt {x^2 + y^2} and \theta = arctan \left( \frac {y}{x} \right).

    In fact

    \theta = arctan \left( \frac {y}{x} \right) + 2k\pi

    for any integer k. You have implicitly chosen k = 0 (and x \ne 0). In doing so you have chosen what is called a "branch" of the log function.

    This is related to the fact that the \log(z) function should be the inverse of the e^z function. But e^z only has an inverse when you restrict the domain so it is one-to-one on that domain. Thus for any constant y_0 \in \mathbb{R} and domain

    D = \{ z = x + iy \in \mathbb{C} | x \in \mathbb{R},\ y \in [y_0, y_0 +2\pi) \},

    the function

    e: D \to \mathbb{C}\setminus\{0\}

    is one-to-one and onto and thus has an inverse

    \log: \mathbb{C}\setminus\{0\} \to D.

    Whereas

    e: \mathbb{C} \to \mathbb{C}\setminus\{0\}

    is not one-to-one and thus does not have an inverse.

    Unfortunately it has been a while since I took complex analysis, so I cannot give you a practical example showing the importance of being explicit about the branch of the log function you chose. I think you just need to be aware that when you use the log function, you have at least implicitly chosen a branch.
    Last edited by JakeD; July 1st 2007 at 12:35 AM.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JakeD View Post
    First, there is a typo:

    \boxed { \log_a (x + iy) = \log_a r + i \pi \log_a e }

    should be

    \boxed { \log_a (x + iy) = \log_a r + i \theta \log_a e }.
    Fixed it, thanks

    Second and more importantly, the \theta is not uniquely defined in

     \boxed { x + iy = r \cdot e^{i \theta} }

    where r = \sqrt {x^2 + y^2} and \theta = arctan \left( \frac {y}{x} \right).

    In fact

    \theta = arctan \left( \frac {y}{x} \right) + 2k\pi

    for any integer k. You have implicitly chosen k = 0 (and x \ne 0). In doing so you have chosen what is called a "branch" of the log function.
    i thought something like + 2k \pi should be there but never thought it was that important, so I just shrugged it off.

    How do you know what branch to choose? That is, how do we know what k we need?

    This is related to the fact that the \log(z) function should be the inverse of the e^z function. But e^z only has an inverse when you restrict the domain so it is one-to-one on that domain.
    now we're really getting into unfamiliar territory. I have zero experience with complex analysis, so a lot of what you're saying is news to me. am i to understand that e^z is not necessarily a one-to-one function? i'm going to have to play around with it to see how that can be algebraically (would i need a lot of complex analysis to show that?) Can we graph such a function?


    Thus for any constant y_0 \in \mathbb{R} and domain

    D = \{ z = x + iy \in \mathbb{C} | x \in \mathbb{R},\ y \in [y_0, y_0 +2\pi) \},

    the function

    e: D \to \mathbb{C}\setminus\{0\}

    is one-to-one and onto and thus has an inverse

    \log: \mathbb{C}\setminus\{0\} \to D.

    Whereas

    e: \mathbb{C} \to \mathbb{C}\setminus\{0\}

    is not one-to-one and thus does not have an inverse.
    Ok, so i was almost completely lost through all of that. it's weird, it's like i understand what you wrote but don't get the meaning behind it. maybe it's because it's almost 4am. I should get some sleep and read it tomorrow.

    Unfortunately it has been a while since I took complex analysis, so I cannot give you a practical example showing the importance of being explicit about the branch of the log function you chose. I think you just need to be aware that when you use the log function, you have at least implicitly chosen a branch.
    Okay, i got that much. Now i just need to figure out which branch to choose... and figure out what exactly is a branch

    Thank you for your response!
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    Quote Originally Posted by Jhevon View Post
    i thought something like + 2k \pi should be there but never thought it was that important, so I just shrugged it off.

    How do you know what branch to choose? That is, how do we know what k we need?
    You can choose whatever k you want.

    now we're really getting into unfamiliar territory. I have zero experience with complex analysis, so a lot of what you're saying is news to me. am i to understand that e^z is not necessarily a one-to-one function? i'm going to have to play around with it to see how that can be algebraically (would i need a lot of complex analysis to show that?) Can we graph such a function?
    Take as an example that e ^{i2k\pi} = 1 for any integer k, which follows from taking the 2k power of Euler's Identity. Thus e^z is many-to-one and defining \log(1) = i 2k\pi for any integer k is possible.

    Now i just need to figure out which branch to choose... and figure out what exactly is a branch
    My complex analysis book defines a branch like this.

    The function \log: \mathbb{C}\setminus\{0\} \to \mathbb{C}, with range y_0 \le \text{Im} \log z < y_0 + 2\pi, is defined by \log z = \log |z| + i \arg z, where \arg z takes values in [y_0,y_0 + 2\pi). Here \log |z| is the usual logarithm of the positive real number |z|. This function is sometimes referred to as the "branch of the logarithm function lying in \{ x+iy\ |\ y_0 \le y < y_0 + 2\pi \}."
    Last edited by JakeD; July 1st 2007 at 02:38 AM.
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    Complex Logarithms are really "complex" are they not as simple as with Real variables. I did not not read everything what JakeD posted so I might be repetative.

    Given the function w=e^z it is natural to define the logarithm function to be the solutions to e^w = z. But the problem is if z_0 is such a number, i.e. e^{z_0}=z then z_0+2\pi i n are too, by Euler's formula because \sin (x+2\pi n)=\sin x and \cos (x+2\pi n)=\cos x.

    So the function f(z)=e^z is not one-to-one. However we can still define a logarithm function at least partially. The typical way we do this is by defining \log z = \ln |z| + i\arg z. Let me explain everything. I use \log to stand for "complex logarithm" and I use \ln to stand for "natural logarithm" (but that is just how I like to write it). Now |z|= \sqrt{x^2+y^2} is the absolute values of the complex number. And \arg z is the "argument" of the complex number. The argument is the angle (-\pi,\pi] the complex number makes with the x-axis (if you think of this geometrically in the Argand diagram). Note this definition is not defined for z=0, that is the single "bad point".

    More imfromation here.

    So I will use \log for the "principal logarithm". And \mbox{Log} for the multi-valued function logarithm, defined to be all the values.

    Example: Given z=1+i we see that \log (1+i) = \ln |1+i| + i \arg (1+i) = \ln \sqrt{2} + \frac{\pi}{4} i. And \mbox{Log} (1+i) = \{ \log (1+i) + 2\pi i n \}.

    Properties of Complex Logarith
    The function f(z) = \ln z is continous everywhere on \mathbb{C} except for the half-line (-\infty,0], this is referred to its "branch". Furthermore this function is holomorphic (which is the complex version of saying "differenciable") on this set as well. Just like with real variables f'(z) = \frac{1}{z} for these differenciable points.

    Why is \log z not continous on that line? I leave that to you to figure out. Take for instance x=-1 and approach that point as a limit first from the top and then from the bottom and see that the values fail to agree. That will imply it is not continous.

    The above shows there are infinitely many versions of logarithm function, because we can chose \arg to represent the numbers [0,2\pi) if we wanted. Or [-\pi/2,3\pi/2) and so on.

    Warning:
    The rules of complex numbers \log (xy) = \log x + \log y might fail.


    Another Version of Complex Logarithm:
    There is another way to define the complex logarithm function.

    I will state this without proof.

    Theorem: Let f(z)=e^z be a complex function defined on an open non-empty set S not containing the origin. Then there exists a holomorphic function g(z) on S such that e^{g(z)} = z for all z\in S.

    That plays the role of logarithm. This function can be completely different from the one I was discussing above, namely because it is holomorphic on its domain. Sometimes we use the notation \log_S z to define this function. So it depends on the S.

    Again, the sad news are there infinitely many such choices for \log_S z and they also differ by 2\pi n i.

    So in the world of complex analysis some things get more difficult.

    ------
    Remark: I do not like to write \sqrt{-1}=i because the function f(z)=\sqrt{z} is again, a multivalued function. Instead it is appropriate to write i^2=-1 because here all you are defining is the binary operation on it self.
    Last edited by ThePerfectHacker; July 1st 2007 at 07:25 AM.
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    A bit of trivia that may amuse you . . . or not.


    Since e^{\frac{\pi}{2}i} \;=\;\cos\!\left(\frac{\pi}{2}\right)+ i\sin\!\left(\frac{\pi}{2}\right) \;=\;i

    . . we have: . \ln(i) \:=\:\frac{\pi}{2}i


    Multiply both sides by i\!:\;\;i\!\cdot\!\ln(i) \;=\;i\!\left(\frac{\pi}{2}i\right)\quad\Rightarro  w\quad \ln(i^i) \;=\;-\frac{\pi}{2}

    . . Therefore: . i^i \;=\;e^{-\frac{\pi}{2}} \;=\;0.2078879576...


    So i raised to the power i is a real number.
    . . Oh well, why not?

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    Here is a math "paradox" I invented some time ago to demonstrate a point:

    1^{\pi} = \left( e^{2\pi i} \right)^{\pi} = e^{2\pi^2 i} = \cos \pi^2 + i \sin \pi^2 \not = 1

    ----
    Taking exponents in real analysis is often defined as,
    a^b = e^{b\ln a} for a>0.
    And e^x is simply the inverse of the function \ln x.

    In complex analysis it again might lead to problems. So we usually define it along the principal logarithm.

    So for example, what Soroban did,
    i^i = e^{i \log (i)} ---> By definition.
    But,
    \log i =\ln |i| + i \arg(i) = \frac{\pi}{2}i
    So,
    i^i = e^{-\pi/2}
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    Quote Originally Posted by ThePerfectHacker View Post
    Warning:
    The rules of complex numbers \log (xy) = \log x + \log y might fail.
    But this holds up to the addition of integral multiples of 2\pi i.

    For example, e^{\pi i} = -1 so \log (-1) = \pi i.

    Using the principal branch of the logarithm function,

    \log( -1 \times -1) = \log(1) = 0 \ne \log(-1) + \log(-1) = 2\pi i,

    but 0 \text{ and } 2 \pi i are "equal" up to addition of integral multiples of 2 \pi i.
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    Quote Originally Posted by JakeD View Post
    But this holds up to the addition of integral multiples of 2\pi i.
    This hold true if you use the concept of the multi-valued function I posted in the link above.

    Thus,
    \mbox{Log}(xy) = \mbox{Log}x + \mbox{Log}y

    Where + here means the sum of those two sets.
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  10. #10
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    What is a "branch cut"? My Complex Analysis book fails to give a good explaination. Furthermore, Bak on Complex Analysis just does not mention it in his book. So I will try to explain it my way.

    Basically a "branch" is what needs to be removed to make the function continous. As I explained \log z is continous on \mathbb{C} except on the half-line (-\infty,0].

    Here is another similar example. In real analysis there is no difficutly understanding \sqrt[n]{x}. Just we sometimes need to restrict our attention to x\geq 0 when n is even. But still, it is rather simple to understand.

    Consider \sqrt[4]{1}. It means the solutions to z^4 = 1. With some factoring we get z=+1,-1,i,-i. There are four solutions. In fact the z^n = 1 has exactly n solutions. They are referred to as the "roots of unity" and (just to get off topic a little bit) of extreme important in field theory (more here.).

    Now it would be nice if we can have that \sqrt[n]{z} be a single valued function. In real analysis it is different, we chose the positive one if there is ambiguity. In complex analysis there is a problem, they are not ordered. So what do we do to keep it single valued?
    Well, we can view this as z^{1/n} and apply the concepts of logarithms to get (by definition) z^{1/n} = \exp \left( \frac{\log z}{n} \right) = \exp \left( \frac{\ln |z|}{n} + i \cdot \frac{\arg z}{n} \right) = \exp \left( \frac{\ln |z|}{n} \right) \cdot \exp \left( i\frac{\arg z}{n} \right) = |z|^{1/n} e^{i\arg z / n}.

    So for example, \sqrt{-1} = |-1|^{1/2} e^{i \arg (-1)/2} = e^{i\pi/2} = \cos \frac{\pi}{2} + i \sin \frac{\pi }{2} = i.

    We ask where is f(z) = \sqrt[n]{n} continous? Is is everywhere except on (-\infty,0], namely because it was derived by the complex logarithm function which fails to be continous there.

    So we say the "branch" of \sqrt[n]{z} is also (-\infty,0] because we removed it "cut it" then we are left with a continous functions.

    I hope I can do the next example, I just made it up and it can get messy.

    Example: What is the branch of f(z) = \log \sqrt{z}.

    Solution:
    1)First z\not =0 because then \log is not defined, so it is not continous there.

    2)Remember the rule, the composition of two continous functions is continous. Now \sqrt{z} is continous everywhere except for (-\infty,0] so let us ignore that. Thus, if \log is continous on the range of \sqrt{z} then by what was said above we have that \log \sqrt{z}. Well is it? Is \log continous on the range of \sqrt{z}.

    I have to stop here. Maybe continous tomorrow. See if you can finish that problem, I think it actually is a hard one.
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Hey guys, I've been busy with some other things so i haven't been able to respond to the posts made. I will soon. I have to put a lot of thought into what i'm going to say because i didn't know what i was getting myself into, complex logarithms and complex exponents is much more "complex" than i thought it was. Anyway, I just wanted to attempt to answer one of the questions posed by TPH. Am i correct?
    Quote Originally Posted by ThePerfectHacker View Post
    (1 + i)^i
    (1 + i)^i = e^{i \ln (1 + i)}

    I found what was \ln (1 + i) in my first post, so i will just plug in the answer if you don't mind.

    \Rightarrow (1 + i)^i = e^{i \left( \ln 2 / 2 + i \pi / 4  \right)}

    = e^{- \pi / 4 + i \ln 2 / 2}

    = e^{- \frac { \pi}{4}} \cdot e^{ i \ln 2 / 2}

    = e^{ - \frac { \pi}{4}} \cdot \left( \cos \left( \frac { \ln 2}{2} \right) + i \sin \left( \frac { \ln 2}{2} \right) \right)

    And i don't want to simplify any further. But am i correct so far?
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  12. #12
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    Yes, that looks correct.
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  13. #13
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    Let me return to the example about finding the branches of f(z) = \ln \sqrt{z}.

    Goal: Find what needs to be removed to make f(z) continous.

    1)This is undefined for z=0, so from now on we will omit this value.

    2)The composition of two continous functions is a continous function (abusing this slightly). That means if we can find a place in \mathbb{C} where \sqrt{z} is continous and \ln z is continous on the range of \sqrt{z} then the composition \ln \sqrt{z} shall be a continous function.

    3)The place where \sqrt{z} is continous is on \mathbb{C} - (-\infty,0]. So if we can find all z on that set so that \sqrt{z} \not \in (-\infty,0] (that is the discontinuity of \ln z) then \ln \sqrt{z} shall be continous by #2.

    4)By #3 we are working on \mathbb{C} - (-\infty, 0]. Now we ask for what of those values is \sqrt{z} \not \in (-\infty,0]. Instead, find the values such that \sqrt{z} \in (-\infty,0] and then remove them. By definition \sqrt{z} = |z|^{1/2} e^{i\cdot \arg z/2}. So if we can this to be on (-\infty,0] we want the real part to be negative and imaginary part to be zero. That is, |z|^{1/2} \cos (\arg z / 2) < 0 \mbox{ and }|z|^{1/2} \sin (\arg z / 2) = 0. Let us look at the second condition: |z|^{1/2} \sin (\arg z/2) = 0. Since we initially set z\not = 0 it must mean that \sin (\arg z /2) = 0 iff \arg z = 2\pi n. Since \arg z \in (-\pi,\pi] it must mean that \arg z = 0 . The only such numbers are the positive numbers, but when we take their square root we still have a positive number, so the second condition |z|^{1/2} \cos (\arg z/2)=0 is impossible.

    We have shown that (-\infty,0] is a branch cut.

    Try to find the branch of \sqrt{\ln z}.
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