Principal branch of the log function

[Measle]

I'm learning complex analysis right now, and I'm reading from Joseph Taylor's Complex Variables.

On Theorem 1.4.8, it says "If a log is the branch of the log function determined by an interval I, then log agrees with the ordinary natural log function on the positive real numbers if and only if the interval I contains 0."

Can anyone help me understand what this means? Is it just saying that the properties of the real log function only hold if the interval I contains 0? I don't understand why.

 

[fresh_42]

(from: https://de.wikipedia.org/wiki/Logarithmus#Komplexer_Logarithmus; engl.: https://en.wikipedia.org/wiki/Logarithm#Complex_logarithm)

You see, that the logarithm $w$ for a value $z$, i.e. $e^w=z$ has all solutions $w+2k\pi i\; , \;k\in\mathbb{Z}$. To make it unique, we have to choose one of these branches, i.e. one value for $k$. This prinicpal value is usually determined by $- \pi < \operatorname{Im}(w) \leq \pi\,.$ Of course we could also say $-\pi/2 < \operatorname{Im}(w) \leq 3/2 \pi$ or any other interval with $0$ in it. But if we climb up, such that zero isn't part of the interval anymore, then we are in the complex world, i.e. we ran around the circle once before regarding the logarithm, but this "run around once" has no real counterpart, because in the reals we would arrive at where we started, whereas in the complexes we climbed up a branch, which is not identical to where we started.

 

[math771]

Hi there! I'm also currently studying complex analysis and I can try to help clarify this theorem for you.

Essentially, what this theorem is saying is that for the complex logarithm function to agree with the real natural logarithm function on positive real numbers, the interval I that determines the branch of the log function must contain 0.

To understand why, let's first remember that the complex logarithm function is defined as log(z) = ln|z| + i arg(z), where arg(z) is the principal argument of z. This means that the complex logarithm function not only takes into account the magnitude of z, but also its direction or argument.

Now, let's consider the real natural logarithm function, which is simply ln(x). This function only takes into account the magnitude of x, not its direction or argument. So, if we want the complex logarithm function to agree with the real natural logarithm function on positive real numbers, the interval I must contain 0.

Why is this the case? Well, if the interval I contains 0, then the principal argument of z will also be 0 for all positive real numbers. This means that the imaginary part of log(z) will be 0, making it equivalent to ln(x). However, if the interval I does not contain 0, then the principal argument of z will vary and the imaginary part of log(z) will not be 0, making it different from ln(x).

I hope this explanation helps! Let me know if you have any other questions. Good luck with your studies!