Sonia Gay

Answered

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:

${\mathrm{log}}_{b}\left({a}^{d}\right)=d{\mathrm{log}}_{b}(a)$

${\mathrm{log}}_{b}(a)=\frac{{\mathrm{log}}_{c}(a)}{{\mathrm{log}}_{c}(b)}$

${\mathrm{log}}_{b}(xy)={\mathrm{log}}_{b}(x)+{\mathrm{log}}_{b}(y)$

${\mathrm{log}}_{b}\left(\frac{x}{y}\right)={\mathrm{log}}_{b}(x)-{\mathrm{log}}_{b}(y)$

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

Are the logarithm rules true for complex numbers?

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

${\mathrm{log}}_{b}\left({a}^{d}\right)=d{\mathrm{log}}_{b}(a)$

${\mathrm{log}}_{b}(a)=\frac{{\mathrm{log}}_{c}(a)}{{\mathrm{log}}_{c}(b)}$

${\mathrm{log}}_{b}(xy)={\mathrm{log}}_{b}(x)+{\mathrm{log}}_{b}(y)$

${\mathrm{log}}_{b}\left(\frac{x}{y}\right)={\mathrm{log}}_{b}(x)-{\mathrm{log}}_{b}(y)$

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

Answer & Explanation

Anika Stevenson

Expert

2022-06-25Added 19 answers

You have to be careful because logs and non-integer powers are multivalued functions. The definition is that ${a}^{d}=\mathrm{exp}(d\mathrm{ln}(a))$ (for any branch of $\mathrm{ln}$). Now ${\mathrm{log}}_{b}({a}^{d})$ is any $z$ such that ${b}^{z}={a}^{d}$, i.e. $\mathrm{exp}(z\mathrm{ln}(b))=\mathrm{exp}(d\mathrm{ln}(a))$, and that is equivalent to $z\mathrm{ln}(b)-d\mathrm{ln}(a)=2\pi in$ for some integer $n$. So the result is

${\mathrm{log}}_{b}({a}^{d})={\displaystyle \frac{d\mathrm{ln}(a)+2\pi in}{\mathrm{ln}(b)}}$

Similarly,

${\mathrm{log}}_{b}(a)={\displaystyle \frac{\mathrm{ln}(a)+2\pi im}{\mathrm{ln}(b)}}$

for some integer $m$. And thus (assuming you use the same values of $\mathrm{ln}(a)$ and $\mathrm{ln}(b)$ in both cases)

${\mathrm{log}}_{b}({a}^{d})-d{\mathrm{log}}_{b}(a)=2\pi i{\displaystyle \frac{n-md}{\mathrm{ln}(b)}}$

For example, take $a=b=e$ and $d=2\pi i$, and use the principal branch of $\mathrm{ln}$

${\mathrm{log}}_{e}({e}^{2\pi i})=\mathrm{ln}(1)=0$ but $2\pi i{\mathrm{log}}_{e}(e)=2\pi i$

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

${\mathrm{log}}_{b}({a}^{d})={\displaystyle \frac{d\mathrm{ln}(a)+2\pi in}{\mathrm{ln}(b)}}$

Similarly,

${\mathrm{log}}_{b}(a)={\displaystyle \frac{\mathrm{ln}(a)+2\pi im}{\mathrm{ln}(b)}}$

for some integer $m$. And thus (assuming you use the same values of $\mathrm{ln}(a)$ and $\mathrm{ln}(b)$ in both cases)

${\mathrm{log}}_{b}({a}^{d})-d{\mathrm{log}}_{b}(a)=2\pi i{\displaystyle \frac{n-md}{\mathrm{ln}(b)}}$

For example, take $a=b=e$ and $d=2\pi i$, and use the principal branch of $\mathrm{ln}$

${\mathrm{log}}_{e}({e}^{2\pi i})=\mathrm{ln}(1)=0$ but $2\pi i{\mathrm{log}}_{e}(e)=2\pi i$

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

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