Complex logarithm

If \(e^{w}=z\) with \(w=u+iv\), then \(e^{u}=|z|\) and \(v\) is an argument of \(z\). Hence \(u=\ln|z|\) and \(v=\arg z+2\pi k\), which gives \(\log z=\ln|z|+i(\arg z+2\pi k)\). The logarithms of \(z\) thus form a family differing by integer multiples of \(2\pi i\), and \(\log 0\) does not exist because \(e^{w}\neq0\).

To get a function one fixes a branch. The most common is the principal logarithm, \(\operatorname{Log}z=\ln|z|+i\operatorname{Arg}z\), with \(\operatorname{Arg}z\in(-\pi,\pi]\). For instance \(\operatorname{Log}i=\tfrac{\pi}{2}i\), \(\operatorname{Log}(-1)=\pi i\) and \(\operatorname{Log}(1+i)=\tfrac12\ln2+\tfrac{\pi}{4}i\). Every logarithm of \(z\) is \(\operatorname{Log}z+2\pi i k\).

This choice has a price: \(\operatorname{Log}\) is discontinuous on the negative reals, where the argument jumps from \(\pi\) to \(-\pi\), and continuous on the rest, \(\mathbb{C}\) minus \((-\infty,0]\). Also, \(e^{\operatorname{Log}z}=z\) always, but \(\operatorname{Log}e^{z}=z\) only if \(\operatorname{Im}z\in(-\pi,\pi]\); for example \(\operatorname{Log}e^{4i}=(4-2\pi)i\).

The rules of the real logarithm hold only up to multiples of \(2\pi i\): \(\log(zw)=\log z+\log w\) as an equality of sets, but not for the principal logarithm (with \(z=w=-1\) it fails by \(2\pi i\)). It is the most common mistake when moving from the real to the complex case.