An angle for $z$ is not unique: adding a full turn $2\pi$ points the same way. Exponential form hid this, because $e^{i\theta} = e^{i(\theta + 2\pi)}$. Before the complex logarithm arrives, we need language for the whole family of angles and a rule for choosing one of them.
🔑 Key idea
For $z \ne 0$, $\arg z$ is the set of all arguments $\theta + 2\pi k$ ($k$ any integer). The principal argument $\operatorname{Arg} z$ is the one value in $(-\pi, \pi]$. Products add arguments as sets, $\arg(zw) = \arg z + \arg w$, but the principal values only satisfy
$$\operatorname{Arg}(zw) = \operatorname{Arg} z + \operatorname{Arg} w + 2\pi k, \qquad k \in \{-1, 0, 1\},$$
with $k \ne 0$ exactly when the sum leaves $(-\pi, \pi]$.
💡 Why it works
If $z = |z|e^{i\alpha}$ and $w = |w|e^{i\beta}$, then $zw = |zw|e^{i(\alpha + \beta)}$, so $\alpha + \beta$ is an argument of $zw$. When $\alpha = \operatorname{Arg} z$ and $\beta = \operatorname{Arg} w$, the sum lies in $(-2\pi, 2\pi]$, which is wider than the principal interval. Shifting it by one full turn (if needed) brings it back. The interval $(-\pi, \pi]$ is a convention: it puts the jump on the negative real axis, and it includes $\pi$ so that negative real numbers have $\operatorname{Arg} = \pi$.
Why care? A function like $\operatorname{Arg}$ has to give one answer per input, so it must make a choice, and every choice has a place where it jumps. For $\operatorname{Arg}$ the jump is on the negative real axis: just above it the value is close to $\pi$, just below it close to $-\pi$. Near that ray, $\operatorname{Arg}$ is not continuous. When we define the complex logarithm by $\operatorname{Log} z = \ln|z| + i\operatorname{Arg} z$, it inherits exactly this jump, and the rules for logarithms inherit the extra $2\pi$.
Drag $z$ around the origin, or press "Go around once". The gold spiral is a continuous argument, which grows by $2\pi$ each lap. The principal value jumps by $2\pi$ when $z$ crosses the red ray, the negative real axis.
Worked examples
✏️ Worked example: Principal arguments on the axes
Find $\operatorname{Arg}(-4)$, $\operatorname{Arg}(-2i)$ and $\operatorname{Arg}(-\sqrt{3} + i)$.
$-4$ lies on the negative real axis: $\operatorname{Arg}(-4) = \pi$ (not $-\pi$, which is excluded).
$-\sqrt{3} + i$ is in quadrant II with reference angle $\tan^{-1}\frac{1}{\sqrt{3}} = \frac{\pi}{6}$, so $\operatorname{Arg} = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.
Answer: $\pi$, $-\frac{\pi}{2}$, $\frac{5\pi}{6}$
✏️ Worked example: When the sum leaves the interval
Let $z = -1 + i$ and $w = -\sqrt{3} + i$. Find $\operatorname{Arg} z + \operatorname{Arg} w$ and $\operatorname{Arg}(zw)$.
$\operatorname{Arg} z = \frac{3\pi}{4}$ and $\operatorname{Arg} w = \frac{5\pi}{6}$; their sum is $\frac{9\pi + 10\pi}{12} = \frac{19\pi}{12}$.
Check: $zw = (-1 + i)(-\sqrt{3} + i) = (\sqrt{3} - 1) - (1 + \sqrt{3})i$, which has a positive real part and a negative imaginary part, so its angle is indeed between $-\frac{\pi}{2}$ and $0$.
Answer: sum $\frac{19\pi}{12}$; $\operatorname{Arg}(zw) = -\frac{5\pi}{12}$
Drag $z$ and $w$ into the left half-plane. The readout compares $\operatorname{Arg} z + \operatorname{Arg} w$ with $\operatorname{Arg}(zw)$ and flags the $2\pi$ difference.
⚠️ Common mistake
Assuming $\operatorname{Arg}(zw) = \operatorname{Arg} z + \operatorname{Arg} w$. For $z = w = -1$, the right side is $\pi + \pi = 2\pi$, but $zw = 1$ and $\operatorname{Arg} 1 = 0$. The set identity $\arg(zw) = \arg z + \arg w$ is fine; the principal values can differ by $2\pi$. Check any answer by asking whether it lies in $(-\pi, \pi]$.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.