What is $\log(z_1z_2)−\log(z_1)−\log(z_2)$ for $z_1$, $z_2$ in the second quadrant?
For the first part of the question $\log(z_1z_2) = \log(z1)+\log(z_2)$, I think I solved it correctly.
By using principle branch, $\log z = \ln z + i\text{Arg}z , (-pi ,pi)$.
$\log(z_1z_2)= \log (z_1z_2) + i( \theta_1 + \theta _2) \dots = \log z_1 + \log z_2 $
Best Answer
If we define the complex logarithm on the principal branch, then
$$\log(z)=\text{Log}(|z|)+i\text{Arg}(z)$$
with $-\pi<\text{Arg}(z)\le \pi$ and $\text{Log}(x)$ denotes the logarithm from real analysis.
If $\pi/2<\text{Arg}(z_1)<\pi$ and $\pi/2<\text{Arg}(z_2)<\pi$, then clearly $\pi<\text{Arg}(z_1)+\text{Arg}(z_2)<2\pi$.
But the principal angle associated with $z_1z_2$ is such that $-\pi<\text{Arg}(z_1z_2)<0$ and we find that $\text{Arg}(z_1)+\text{Arg}(z_2)=\text{Arg}(z_1z_2)+2\pi$.
We conclude, therefore, that
$$\begin{align} \log(z_1z_2)&=\text{Log}(|z_1z_2|)+i\text{Arg}(z_1z_2)\\\\ &=\text{Log}(|z_1|)+\text{Log}(|z_2|)+\text{Arg}(z_1)+\text{Arg}(z_2)-2\pi\\\\ &=\log(z_1)+\log(z_2)-2\pi \end{align}$$
And we are done.