Good day everyone. I was reading the more advanced lectures on complex analysis and encountered a lot of questions, concerning the determination of complex logarithm. As far I don't even understand the concept of it, but I'll do provide you with several practical questions concerning the topic.

First of all, the determination of complex logarithm on open $\Omega \subset \mathbb{C}$ is continuous $f(w)$ such that:

$$\forall w \in \Omega \text{ }\exp(f(w))=w$$

So then it starts. Maybe someone could explain what presumptions does the following statement contradicts.

There is no continuous determination for complex logarithm in $\mathbb{C}$ \ ${0}$.

Second part is more practical.

We say that determination of complex logarithm is called principal if it given as a complement of $\mathbb{C}$ and semi axis of negative or zero reals $\Omega_{\pi}=\mathbb{C}$ \ $\{z\in \mathbb{C} : \Re(z) \leq 0\}$. Such as $$ f(z) = \log(|z|)+ i\begin{cases} \arcsin{(y/|z|)} & x \geq 0, \\ \pi - \arcsin{(y/|z|)} & x \leq 0,\, y \geq 0, \\ -\pi - \arcsin{(y/|z|)} & x \leq 0,\, y \leq 0. \end{cases}$$

We can see that the argument belongs to $(-\pi,\pi]$, but I don't understand neither why a set without a negative reals define such argument neither how does it happens. After this there is an example saying that if we take $\Omega_0=\mathbb{C}$ \ $\{z \in \mathbb{C} : \Re \geq 0\}$ then the argument will be $(0,2\pi]$ but I also didn't get how this happens. What will happen if we take out non positive imaginary part? What kind of argument we will have then?

There is no such topic explained in wiki so maybe the deep answer will help others who encountered the same problem.

  • $\begingroup$ "Maybe someone would explain what presumptions does the following statement contradicts" is unclear -- could you perhaps rephrase that? (Also the singular of mathematical "axes" is "axis"; the singular is "axe" only when "axes" refers to the instruments used to chop down trees.) $\endgroup$ – joriki Apr 25 '12 at 3:16
  • $\begingroup$ "Maybe someone would explain what presumptions does the following statement contradicts" I meant at what point do we found our statement to be incorrect or simply where does it goes wrong. I am sorry for grammatical mistakes it's the problem of multiple translations. $\endgroup$ – user974514 Apr 25 '12 at 11:36
  • 1
    $\begingroup$ The statement is correct. It seems that what you mean is not where the statement goes wrong but where an attempt at constructing a continuous logarithm on $\mathbb C\setminus\{0\}$ goes wrong? $\endgroup$ – joriki Apr 25 '12 at 11:45

By inverting Euler's formula $e^{ix} = \cos{x} + i \sin{x}$, you get

$\log(r(\cos{\theta} + i \sin{\theta}))$

$ = \log{r} + \log(\cos{\theta} + i \sin{\theta})$

$ = \log{r} + i \theta$

The problem here is that $\theta$, the argument or phase angle of a complex number, is not unique, due to sin and cos being periodic. For example, the following are all valid polar representations of $-1-i$:

  • $\sqrt{2} \text{ cis } \frac{-3\pi}{4}$
  • $\sqrt{2} \text{ cis } \frac{5\pi}{4}$
  • $\sqrt{2} \text{ cis } \frac{-11\pi}{4}$

So, how do you decide whether $\log (-1-i)$ is $\frac{\log{2}}{2} - \frac{3\pi}{4} i$, $\frac{\log{2}}{2} + \frac{5\pi}{4} i$, or $\frac{\log{2}}{2} - \frac{11\pi}{4} i$? By arbitrary picking a $2\pi$-wide interval to which $\theta$ is restricted.

  • If you choose $\theta \in (-\pi, \pi]$, then $\log (-1-i) = \frac{\log{2}}{2} - \frac{3\pi}{4} i$.
  • If you choose $\theta \in [0, 2\pi)$, then $\log (-1-i) = \frac{\log{2}}{2} + \frac{5\pi}{4} i$.

Whichever "branch cut" you choose, there will be a discontinuity across it. For example, using $\theta \in (-\pi, \pi]$, then if $x$ is a negative real number,

  • $\lim_{y \rightarrow 0^+} (x + yi) = \log{|x|} + \pi i$
  • $\lim_{y \rightarrow 0^-} (x + yi) = \log{|x|} - \pi i$

I think your first question is, why is there no continuous logarithm function defined in the complex plane, omitting $0$. The answer is, trace out a circle of radius $1$ centered at the origin, starting at $1$, and evaluate the logarithm function as you go around. Go ahead, do it! You will find that you either make a discontinuity somewhere along the way, or else when you get back to $z=1$ you are off by $2\pi i$ from the value you started with.

Now the same thing happens with the argument function. Start it with, say, the value $0$ at $z=1$, go around that circle (counterclockwise, just to be specific), and when you come back to $z=1$ the argument will be $2\pi$; discontinuity! The cure for the discontinuity is to make it impossible to go all the way around the circle. The easiest way to do that is to remove some ray that starts at the origin and goes to infinity. If the ray you remove is the non-positive reals, then the argument, starting with the value $0$ at $z=1$, will increase almost but not quite to $\pi$ as you go around counterclockwise (not reaching the excised ray), and will decrease almost but not quite to $-\pi$ as you go around the other way.

The non-positive imaginary part ray has argument $3\pi/2$ as you approach it counterclockwise from $z=1$, and argument $-\pi/2$ as you approach it the other way, so if you omit that ray then the argument can be taken to be between $-\pi/2$ and $3\pi/2$.

If I have missed the point of your question, please try again.


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