What does $\ln{z} - \ln{z}$ equal, given $z \in \mathbb{C}$?

Question

On first glance, the expression $$ \ln{z} - \ln{z} $$ where $z$ is complex and of form $a + bi$ should always evaluate to zero. Subtracting something by itself should be zero. However, when one takes the multivaluedness of the complex logarithm into account, one might find that $$ \frac{1}{2}\ln(a^2+b^2)+i(Arg(z) + 2\pi k_1)\\ - \frac{1}{2}\ln(a^2+b^2)-i(Arg(z) + 2\pi k_2) \\ =2 \pi i(k_1-k_2) $$ where $k_1,k_2 \in \mathbb{Z}$. Let $c = k_1-k_2$ and we finally have the relation $$ \ln{z} - \ln{z} = 2 \pi i c $$ where $c \in \mathbb{Z}$. This feels incredibly hacky and not at all rigorous. However, I can't quite see where my problem lies. Is multivaluedness not treated this way with rigor? Which step did I commit some "division by zero" type error? Or if this is true, is there an intuition for it?

Answer 1

No, $\ln(z)-\ln(z)=0$ and $k_1=k_2$ in what you wrote. When you write $\ln(z)=\ln|z|+i\arg(z)$, then $\arg(z)\in (-\pi,\pi]$. If you don't do such a restriction, then $\arg(z)$ is not a function.