Using the definition of the principal logarithm, investigate…

Using the definition of the principal logarithm, investigate the identity $\ \text{Ln}(z_1z_2)=\text{Ln}\,z_1+\text{Ln}\,z_2$.

Step 1 · Compare both sides

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

$$ \text{Ln}(z_1z_2)=\ln|z_1|+\ln|z_2|+i\,\text{Arg}(z_1z_2), \qquad \text{Ln}\,z_1+\text{Ln}\,z_2=\ln|z_1|+\ln|z_2|+i(\theta_1+\theta_2) $$

The moduli always agree; equality of the two full expressions reduces to whether $\text{Arg}(z_1z_2)=\theta_1+\theta_2$.

Step 2 · When it holds

If $\theta_1+\theta_2\in(-\pi,\pi]$, then this value already is the principal argument of $z_1z_2$, so $\text{Arg}(z_1z_2)=\theta_1+\theta_2$ and the identity holds exactly.

Step 3 · When it fails

If $\theta_1+\theta_2$ falls outside $(-\pi,\pi]$, the principal argument must be brought back into range by adding or subtracting $2\pi$, so:

$$ \text{Ln}(z_1z_2)=\text{Ln}\,z_1+\text{Ln}\,z_2+2\pi ik, \qquad k\in\{-1,0,1\} $$

Step 4 · Counterexample

Take $z_1=z_2=-1=e^{i\pi}$, so $\theta_1=\theta_2=\pi$ and $\theta_1+\theta_2=2\pi\notin(-\pi,\pi]$:

$$ \text{Ln}(-1)=i\pi \ \Longrightarrow\ \text{Ln}\,z_1+\text{Ln}\,z_2=i\pi+i\pi=2\pi i $$

But $z_1z_2=(-1)(-1)=1$, so $\text{Ln}(z_1z_2)=\text{Ln}(1)=0\neq2\pi i$.

Verification

Direct check: $0=2\pi i-2\pi i$, matching the correction formula with $k=-1$.