Are the logarithm rules true for complex numbers?

We know that for positive real numbers $a$, $b$, $c$ and real number $d$ that:

$$\log_b\left(a^d\right)=d\log_b(a)$$ $$\log_b(a) = \frac{\log_c(a)}{\log_c(b)}$$ $$\log_b(xy)=\log_b(x)+\log_b(y)$$ $$\log_b\left(\frac xy\right)=\log_b(x)-\log_b(y)$$

We also know that


Does this extend to complex numbers as $a$, $b$, $c$, or $d$?

My instinct is that

$$\log_b\left(b^{s+t\mathrm i}\right) = s+t\mathrm i$$

In other words, I'm pretty confident that the last formula works when $d$ is a complex number

  • $\begingroup$ Why not? Until you avoid dividing by 0 and putting base as 0, all formulas should work out in the same way as for real numbers. $\endgroup$ – L__ Feb 20 '14 at 6:45
  • $\begingroup$ There is a complication in that a complex number has an infinite number of logarithms, since $ \ a = r \cdot e^{i \theta} \ . $ So any "angle-name" $ \ \theta + 2k \pi \ $ gives us the same complex number. $\endgroup$ – colormegone Feb 20 '14 at 6:48

You have to be careful because logs and non-integer powers are multivalued functions. The definition is that $a^d = \exp(d \ln(a))$ (for any branch of $\ln$). Now $\log_b(a^d)$ is any $z$ such that $b^z = a^d$, i.e. $\exp(z \ln(b)) = \exp(d \ln(a))$, and that is equivalent to $z \ln(b) - d \ln(a) = 2 \pi i n$ for some integer $n$. So the result is $$ \log_b(a^d) = \dfrac{d \ln(a) + 2 \pi i n}{\ln(b)}$$ Similarly, $$\log_b(a) = \dfrac{\ln(a) + 2 \pi i m}{\ln(b)}$$ for some integer $m$. And thus (assuming you use the same values of $\ln(a)$ and $\ln(b)$ in both cases) $$\log_b(a^d) - d \log_b(a) = 2 \pi i \dfrac{n - m d}{\ln(b)}$$ For example, take $a = b = e$ and $d = 2 \pi i$, and use the principal branch of $\ln$. $\log_e(e^{2\pi i}) = \ln(1) = 0$ but $2 \pi i \log_e(e) = 2 \pi i$.

Another interesting example is $a = b = -1$, $d = 3$. Now $(-1)^3 = -1$, but there is no way to have $\log_{-1}(-1) = \log_{-1}((-1)^3) = 3 \log_{-1}(-1)$ (this would imply $\log_{-1}(-1) = 0$, which is certainly false).


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