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.
Best Answer
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 $$ \log z^n = \log 1 = 0 \neq 2\pi i = n \cdot \frac{2\pi i}{n} = n\log z. $$