Sonia Gay

2022-06-24

Logarithm rules for complex numbers
Are the logarithm rules true for complex numbers?
We know that for positive real numbers $a$, $b$, $c$ and real number $d$ that:
${\mathrm{log}}_{b}\left({a}^{d}\right)=d{\mathrm{log}}_{b}\left(a\right)$
${\mathrm{log}}_{b}\left(a\right)=\frac{{\mathrm{log}}_{c}\left(a\right)}{{\mathrm{log}}_{c}\left(b\right)}$
${\mathrm{log}}_{b}\left(xy\right)={\mathrm{log}}_{b}\left(x\right)+{\mathrm{log}}_{b}\left(y\right)$
${\mathrm{log}}_{b}\left(\frac{x}{y}\right)={\mathrm{log}}_{b}\left(x\right)-{\mathrm{log}}_{b}\left(y\right)$
We also know that
${\mathrm{log}}_{b}\left({b}^{d}\right)=d$
Does this extend to complex numbers as $a$, $b$, $c$ or $d$?
My instinct is that
${\mathrm{log}}_{b}\left({b}^{s+t\mathrm{i}}\right)=s+t\mathrm{i}$
In other words, I'm pretty confident that the last formula works when $d$ is a complex number

Anika Stevenson

Expert

You have to be careful because logs and non-integer powers are multivalued functions. The definition is that ${a}^{d}=\mathrm{exp}\left(d\mathrm{ln}\left(a\right)\right)$ (for any branch of $\mathrm{ln}$). Now ${\mathrm{log}}_{b}\left({a}^{d}\right)$ is any $z$ such that ${b}^{z}={a}^{d}$, i.e. $\mathrm{exp}\left(z\mathrm{ln}\left(b\right)\right)=\mathrm{exp}\left(d\mathrm{ln}\left(a\right)\right)$, and that is equivalent to $z\mathrm{ln}\left(b\right)-d\mathrm{ln}\left(a\right)=2\pi in$ for some integer $n$. So the result is
${\mathrm{log}}_{b}\left({a}^{d}\right)=\frac{d\mathrm{ln}\left(a\right)+2\pi in}{\mathrm{ln}\left(b\right)}$
Similarly,
${\mathrm{log}}_{b}\left(a\right)=\frac{\mathrm{ln}\left(a\right)+2\pi im}{\mathrm{ln}\left(b\right)}$
for some integer $m$. And thus (assuming you use the same values of $\mathrm{ln}\left(a\right)$ and $\mathrm{ln}\left(b\right)$ in both cases)
${\mathrm{log}}_{b}\left({a}^{d}\right)-d{\mathrm{log}}_{b}\left(a\right)=2\pi i\frac{n-md}{\mathrm{ln}\left(b\right)}$
For example, take $a=b=e$ and $d=2\pi i$, and use the principal branch of $\mathrm{ln}$
${\mathrm{log}}_{e}\left({e}^{2\pi i}\right)=\mathrm{ln}\left(1\right)=0$ but $2\pi i{\mathrm{log}}_{e}\left(e\right)=2\pi i$
Another interesting example is $a=b=-1$, $d=3$. Now $\left(-1{\right)}^{3}=-1$, but there is no way to have ${\mathrm{log}}_{-1}\left(-1\right)={\mathrm{log}}_{-1}\left(\left(-1{\right)}^{3}\right)=3{\mathrm{log}}_{-1}\left(-1\right)$ (this would imply ${\mathrm{log}}_{-1}\left(-1\right)=0$, which is certainly false).

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