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Although it must be a silly question, I am really confused.

For complex logarithm, in general, $$\log(z_1z_2)\neq\log(z_1)+\log(z_2)$$ even if the logarithm is already determined. I think this is true, right?

However, in some proof from complex analysis, one uses logarithm to let a product (even infinite product) be a summation without considering any determination of logarithm. Is that correct?

For example:

We know that $$\prod_{p}\left(1-\frac{1}{p^s}\right)^{-1}=\sum_{n=1}^\infty\frac{1}{n^s}=\zeta(s)$$ holds for $Re s>1$. In some books, the author states that by $$\log\zeta(s)=-\sum_p\log\left(1-\frac{1}{p^s}\right)$$

Is that correct?

Sometimes, I think complex logarithm is really "annoying", because I always have to worry about the determination of it. Do I have to worry about that?

Thank you very much!

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up vote 7 down vote accepted

First, note that $\log(xy) = \log(x) + \log(y)$ is true for the multivalued $\log$ function, in the sense that both sides have the same set of values.

If you want equality of a particular branch of the logarithm, it takes more care. In the situation you describe it's pretty simple, though: I'll let you work out the

Theorem: If $\text{Re}(z) > 0$ and $\text{Re}(w) \geq 0$, then $\text{Log}(zw) = \text{Log}(z) + \text{Log}(w)$.

where $\text{Log}$ is the principal branch.

In general, you have to account for the branch cut somehow. One way is to do the calculation with the multi-valued logarithm, and at the end figure out which branch the result is supposed to be on.

Using your example again, the difference between the left and right hand sides is an integer multiple of $2 \pi \mathbf{i}$.

If $s$ is real, then both sides are real, so you know exactly which integer multiple to use (i.e. zero).

Then, as $s$ varies continuously over the set $\text{Re}(s) > 1$, you can check that none of the values on the right hand side pass through the branch cut.

If you can verify the left hand side also doesn't pass through the branch cut, then both sides are equal for all $s$ in that domain. On the other hand, if $\zeta(s)$ does pass through the negative real axis, then you have to tally up a copy of $2 \pi \mathbf{i}$ into the left hand side, depending on which way it passes through the cut.

(I don't know enough about $\zeta$ to know which happens)

For a simpler example, consider the function $f(z) = \text{Log}(z^2) - 2 \text{Log}(z)$ defined for non-zero $z$. Clearly $f(1) = 0$. The branch cuts of this function are the entire imaginary axis and the negative real axis.

So, everywhere in the right half plane, we have $f(z) = 0$. Moving z through the first quadrant into the second, we pass the positive imaginary axis. $z^2$ passes through the branch cut of $\text{Log}$ from top to bottom, and so $\text{Log}(z^2)$ decreases by $2 \pi \mathbf{i}$. Thus $f(z) = -2 \pi \mathbf{i}$ everywhere in the second quadrant.

Moving from the second to the third quadrant, $z$ passes through the branch cut of $\text{Log}$, and so $f(z)$ decreases by $(-2) \cdot (2 \pi \mathbf{i})$, so $f(z) = 2 \pi \mathbf{i}$ in the third quadrant.

Finally, moving from the third to the fourth quadrant, $f(z)$ decreases by $2 \pi \mathbf{i}$ again, thus returning its value to $0$.

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Much like the inverse trig functions, the logarithm may take on an infinite set of values:

$$\log{z} = \log{|z|} + i (\theta + 2 \pi k) $$

where $k \in \mathbb{Z}$ and $\theta = \arg{z}$. In this sense, you are right. Nevertheless, if we define a principal branch of $\log{z}$ such that

$$\arg{z} = \mathrm{Tan}^{-1}{\frac{\Im{z}}{\Re{z}}}$$

where $\mathrm{Tan}^{-1}$ refers to the principal branch of the $\tan^{-1}$ function, with a range of $[-\pi,\pi)$, we may then define a unique value of the logarithm of a complex number. Some use the notation $\mathrm{Log{(z)}}$ to denote this principal value such that $\Im{\mathrm{Log{(z)}}}$ takes on a unique value. In this sense, then, $\mathrm{Log{(z_1 z_2)}} = \mathrm{Log{(z_1)}} + \mathrm{Log{(z_2)}}$.

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This is not the usual definition of the principal branch. (The argument should be in $(-\pi,\pi)$ not $[-\pi/2,\pi/2]$.) – mrf Jan 18 '13 at 10:12
The edit doesn't help (completely): $\arg z$ is not a function of $\Im z/\Re z$, since $1+i$ and $-1-i$ should have different arguments. – mrf Jan 18 '13 at 12:04
This is why I has defined the principal branch as being from $[-\pi/2,\pi/2)$. Let's call our function $ArcTan_2(x,y)$ and define it as we know it, so that it is in (-\pi,\pi)$ and gets the quadrant correct. – Ron Gordon Jan 18 '13 at 12:10
Agreed. $\operatorname{ArcTan}_2$ is a very convenient function. – mrf Jan 18 '13 at 13:32

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