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Let $z_1, \dots ,z_n$ be complex numbers such that $\Re(z_k)>0$ and $\Re(z_1 \dots z_n)>0$ for $1 \leq k \leq n$. Show that

$$\log(z_1 \dots z_n) = \log(z_1)+\dots +\log(z_n)$$

where $\log$ is the principal branch of the logarithm. If the restrictions on the $z_k$ are removed, does the formula remain valid.

My Solution

Let $z_1, \dots ,z_n$ be complex numbers such that $\Re(z_k)>0$ and $\Re(z_1 \dots z_n)>0$ for $1 \leq k \leq n$. Consider

\begin{align*} \log(z_1 \dots z_n)&=\ln(|z_1||z_2|\dots|z_n|)+i\arg(z_1 \dots z_n)\\ &= \ln(|z_1|) + \dots + \ln(|z_n|) + i\arg(z_1)+\dots i\arg(z_n)\\ &=\log(z_1) + \dots + \log(z_n) \end{align*}

Now, clearly we must always exclude any $z_k=0$.

My Question

The log function loses continuity if I include the negative real axis (I'm pretty sure at least.) but I am having trouble determining if the function is still valid if the restriction $\Re(z_k)$ is removed. Thanks for your help!

Edit

As pointed out in the comments my solution is fallacious. I have found an example that shows it fails if $\Re(z)<0$. So, we are assuming that $-\dfrac{\pi}{2}<\arg(z_k)<\dfrac{\pi}{2}$. We need to show that $\arg(z_1 \dots z_n)=\arg(z_1)+ \dots+\arg(z_n)$ NOT $\arg(z_1 \dots z_n)=\arg(z_1)+ \dots+\arg(z_n) \quad \text{ mod }2 \pi$. I am not sure how to go about this from here.

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  • $\begingroup$ I am definitely not familiar with complex logarithm. But my suggestion is: why don't you choose another branch? Since you have only finitely many $z_i$, and none of them equals to $0$, then you can easily pick a $\theta$ s.t. the line $e^{i\theta}$ becomes the branch cut. Or you may just want to claim that there exists a simply connected domain which contains all $z_i$, and does not contains $0$. Then you can still have it holds. You may also want to take $z_{i_i} \dots z_{i_k}$, which is the product of an arbitrary subset of $z_1, \dots, z_n$ into account s.t. they won't fall on your line. $\endgroup$ – k99731 Feb 27 '16 at 2:09
  • $\begingroup$ Did you consider to use the representation $re^{i\theta}$? You will get I believe more than one potential outcome. $\endgroup$ – Moti Feb 27 '16 at 2:55
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The formula is not valid even with the initial restrictions (but I suspect OP has a small typo), and it is definitely not valid with them removed.

The principal branch of the logarithm is defined (on $\mathbb{C}$ minus the origin and negative real axis) as $\log z = \log |z| + i \arg z$, where $\arg z \in (-\pi, \pi)$. The absolute value piece gives us no trouble when calculating the logarithm of a product, but we need to take care with the $\arg$. Applying the definition, $$ \begin{eqnarray} \log (z_1z_2\cdots z_k) &=& \log|z_1z_2\cdots z_k|+i\arg(z_1z_2\cdots z_k)\\ &=&\log|z_1|+\log|z_2|+\cdots+\log|z_k|+i\arg(z_1z_2\cdots z_k)\\ &=& \log z_1 + \log z_2 + \cdots + \log z_k + i\left(\arg \prod_{i=1}^{n}z_i - \sum_{i=1}^{n} \arg z_i\right).\end{eqnarray} $$ The term in parentheses is equal to $2\pi$ times the winding number of the path $1,z_1, z_1z_2, \ldots, z_1 z_2\cdots z_n,1$ around the origin (note that none of the segments can cross the origin, because none of the $z_i$ are on the negative real axis). (Equivalently, it's the signed number of times the path crosses the branch cut.) Now, if we assert that $\Re(z_1 z_2 \cdots z_k) > 0$ for all $1\le k \le n$ (which is what OP's equation should read), then the entire path is confined to the right half-plane and clearly can't circle the origin... in which case the winding number is zero and the usual log formula works. On the other hand, if we remove that restriction for the partial products (even leaving it in place for the individual terms), we can easily find cases where the winding number is non-zero. For instance, take $z_k=z=\exp(2\pi i / n)$ for all $1\le k \le n$ (which has positive real part for $n>4$). Then $$ \log z^n = \log 1 = 0 \neq 2\pi i = n \cdot \frac{2\pi i}{n} = n\log z. $$

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In the first part, you are assuming the very thing you are supposed to prove: How do you know that $\arg(z_1...z_n) = \arg(z_1) + ... + \arg(z_n)$?

As for the second part, consider $\left(\frac{-1 + i}{\sqrt 2}\right)^2 = (e^{i(3\pi/4)})^2$.

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  • $\begingroup$ Because the argument of the product is the sum of the arguments. Let $z_i=r_ie^{i\theta_i}$. $\arg{\Pi z_i}=\sum\arg{z_i}$ $\endgroup$ – Aaron Zolotor Feb 27 '16 at 4:04
  • $\begingroup$ No, that is just you restating your assumption that this holds. You need to prove it. That is the point of this exercise. As as my example shows. It does NOT hold in general, when you restrict $\arg$ to the principle branch. $\endgroup$ – Paul Sinclair Feb 27 '16 at 4:08
  • $\begingroup$ I don't understand. That is just a fact about complex numbers. They can be represented in polar form. We look at the product. The magnitude is the product of the magnitude and the argument is the sum of the arguments? $\endgroup$ – Aaron Zolotor Feb 27 '16 at 4:10
  • $\begingroup$ Try my example. what is $\left(\frac{-1 + i}{\sqrt 2}\right)^2$? What is the argument of this number? Then what is twice the argument of $\frac{-1 + i}{\sqrt 2}$? (I am assuming by "principle branch", you mean arguments between $-\pi$ and $\pi$. If you mean $0$ to $2\pi$ instead, then a different example.is needed.) $\endgroup$ – Paul Sinclair Feb 27 '16 at 4:14
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    $\begingroup$ But you are applying to the logarithm, which is restricted in its codomain. For it to be true, you have to show that the RHS arguments sum up to the particular argument on the LHS, not some equivalent one. $\endgroup$ – Paul Sinclair Feb 27 '16 at 22:54

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