The logarithm should undo the exponential: $\log w$ is a number $z$ with $e^z = w$. But the complex exponential repeats every $2\pi i$, so $e^z = w$ has infinitely many solutions. The logarithm therefore has infinitely many values, one for each choice of argument.
Key idea
For $z \ne 0$,
$$\log z = \ln|z| + i\arg z = \ln|z| + i(\operatorname{Arg} z + 2\pi k), \quad k = 0, \pm 1, \pm 2, \ldots$$
is the set of all logarithms; its values are $2\pi i$ apart. The principal logarithm is the single value $\operatorname{Log} z = \ln|z| + i\operatorname{Arg} z$, with imaginary part in $(-\pi, \pi]$. Always $e^{\log z} = z$, but $\operatorname{Log}(e^z) = z$ only when $-\pi < \operatorname{Im} z \le \pi$, and $\operatorname{Log}(zw)$ can differ from $\operatorname{Log} z + \operatorname{Log} w$ by $\pm 2\pi i$.
Why it works
Write $z = |z|e^{i\theta}$ and look for $w = a + ib$ with $e^w = z$. Then $e^a e^{ib} = |z|e^{i\theta}$: the moduli give $e^a = |z|$, so $a = \ln|z|$, and the angles give $b = \theta + 2\pi k$. Every choice of $k$ works, which is why $\log z$ is a set. Choosing the principal argument picks one value; that choice is what makes $\operatorname{Log}$ a function, and its price is the $2\pi$ correction inherited from $\operatorname{Arg}$.
Worked examples
Worked example: Principal values
Find $\operatorname{Log}(-1)$, $\operatorname{Log}(2i)$ and $\operatorname{Log}(1 - i)$.
- $\operatorname{Log}(-1) = \ln 1 + i\pi = \pi i$.
- $\operatorname{Log}(2i) = \ln 2 + \frac{\pi}{2}i$.
- $|1 - i| = \sqrt{2}$ and $\operatorname{Arg}(1 - i) = -\frac{\pi}{4}$: $\operatorname{Log}(1 - i) = \ln\sqrt{2} - \frac{\pi}{4}i = \frac{1}{2}\ln 2 - \frac{\pi}{4}i$.
Answer: $\pi i$; $\ln 2 + \frac{\pi}{2}i$; $\frac{1}{2}\ln 2 - \frac{\pi}{4}i$
Worked example: All the values
List the values of $\log(-3)$ with imaginary part between $-4\pi$ and $4\pi$.
- $\ln|-3| = \ln 3$, and the arguments of $-3$ are $\pi + 2\pi k$.
- Between $-4\pi$ and $4\pi$: $-3\pi, -\pi, \pi, 3\pi$.
- $\ln 3 - 3\pi i$, $\ln 3 - \pi i$, $\ln 3 + \pi i$, $\ln 3 + 3\pi i$.
Answer: $\ln 3 + (2k + 1)\pi i$ for $k = -2, -1, 0, 1$
Worked example: A product rule that fails
Compare $\operatorname{Log}\big((-1)(-1)\big)$ with $\operatorname{Log}(-1) + \operatorname{Log}(-1)$.
- $(-1)(-1) = 1$ and $\operatorname{Log} 1 = 0$.
- $\operatorname{Log}(-1) + \operatorname{Log}(-1) = \pi i + \pi i = 2\pi i$.
- They differ by $2\pi i$. As sets, $\log(zw) = \log z + \log w$ is true; for principal values it can fail.
Answer: $0$ versus $2\pi i$
For a positive real number $x$, $\operatorname{Log} x = \ln x$, so the principal logarithm extends the familiar one. The new values are the extra imaginary parts: $\log x = \ln x + 2\pi ki$. Off the cut, $\operatorname{Log}$ is analytic with derivative $\frac{1}{z}$ (next lesson), which is what makes it useful in calculus.
Logarithms with other bases follow: $\log_b z = \frac{\log z}{\log b}$ for a chosen value of $\log b$. They are rarely needed in complex analysis, where "log" always means the natural logarithm.
Common mistake
Giving $\log z$ one value, or using $\operatorname{Log}(zw) = \operatorname{Log} z + \operatorname{Log} w$ blindly. Unless a problem says "principal", $\log z$ means the whole set $\ln|z| + i(\theta + 2\pi k)$. And the product rule for $\operatorname{Log}$ is off by $2\pi i$ whenever $\operatorname{Arg} z + \operatorname{Arg} w$ leaves $(-\pi, \pi]$, as with $\operatorname{Log}(-1) + \operatorname{Log}(-1) = 2\pi i \ne 0 = \operatorname{Log} 1$. Check: exponentiate your answer; every correct value of $\log z$ gives back $z$.