Using the definition of the principal logarithm, investigate…

Using the definition of the principal logarithm, investigate the identity $\ \text{Ln}\dfrac{z_1}{z_2}=\text{Ln}\,z_1-\text{Ln}\,z_2$.

Step 1 · Compare both sides

Let $\theta_1=\text{Arg}(z_1),\ \theta_2=\text{Arg}(z_2)$. Then:

$$ \text{Ln}\frac{z_1}{z_2}=\ln|z_1|-\ln|z_2|+i\,\text{Arg}\!\left(\frac{z_1}{z_2}\right), \qquad \text{Ln}\,z_1-\text{Ln}\,z_2=\ln|z_1|-\ln|z_2|+i(\theta_1-\theta_2) $$

Step 2 · When it holds and when it fails

If $\theta_1-\theta_2\in(-\pi,\pi]$, the identity holds exactly. Otherwise the principal argument is shifted by $2\pi$, giving:

$$ \text{Ln}\frac{z_1}{z_2}=\text{Ln}\,z_1-\text{Ln}\,z_2+2\pi ik, \qquad k\in\{-1,0,1\} $$

Step 3 · Counterexample

Take $z_1=1=e^{i\cdot0},\ z_2=-1=e^{i\pi}$, so $\theta_1=0,\ \theta_2=\pi$ and $\theta_1-\theta_2=-\pi$, which lies outside the half-open interval $(-\pi,\pi]$ (the endpoint $-\pi$ is excluded):

$$ \text{Ln}\,z_1-\text{Ln}\,z_2=0-i\pi=-i\pi $$

But $z_1/z_2=1/(-1)=-1$, whose principal argument is $+\pi$ (not $-\pi$, by convention), so:

$$ \text{Ln}\!\left(\frac{z_1}{z_2}\right)=\text{Ln}(-1)=i\pi\neq-i\pi $$

Verification

$i\pi=-i\pi+2\pi i$, matching the correction formula with $k=1$.