Show $\log z_1z_2 \neq \log z_1 + \log z_2$ given $z_1 = i$ and $z_2 = -\sqrt 3 + i$.

Question

Show by evaluating both sides that for $z_1 = i$ and $z_2 = -\sqrt 3 + i$,

$\log z_1z_2 \neq \log z_1 + \log z_2$.

Recall the definition: $$\log z = \log |z| + i\arg z$$

Attempt:

Left side:

$\log z_1z_2 = \log [i (-\sqrt 3 + i)] = \log[ i^2 - \sqrt 3i] = \log |- 1 - \sqrt 3 i| + i\arg(-1 - \sqrt 3i) = \log 2 + i(4\pi/3)$

and right side:

$\log z_1 + \log z_2 = \log i + \log (-\sqrt 3 + i) = {\log|i| + i\arg(i)} + {\log|-\sqrt 3 + i)|} + i\arg(-\sqrt 3 + i) = \log 1 + i (\pi/2) + \log 2 + i(5\pi/6) = \log 2 + i(\pi/2 + 5\pi/6) = \log 2 + i(4\pi/3) $.

But I get the same answer. Can someone please tell where the error is? Any feedback will really help. Thank you.

Answer 1

The $Arg(z)$ function spits out the principle argument, which is in the range $(-\pi,\pi]$. So your answer to the first one is wrong, it should be $\frac {-2\pi}3 i$ as the imaginary part

Answer 2

The principal-branch Arg is always between $-\pi$ and $\pi$, so $\text{Arg}(-1-i\sqrt{3})$ is not $4\pi/3$.