2
$\begingroup$

Consider the following two expressions

\begin{equation} \left(\frac{a}{b}\right)^c \end{equation} \begin{equation} \frac{a^c}{b^c} \end{equation}

where $a,b,c$ are complex numbers.

I would like to know for what values of $a,b,c$ are the two expressions equivalent.

One example where they aren't equivalent is $b = c = 0$, then the first expression is undefined and the second expression is $1$. It's also clear that equivalence holds if $b \neq 0$ and $c \in \mathbb{Z}$. Could somebody help me determine the complete set of values where equivalence holds?

P.S. This question is derived from the SO question here.

$\endgroup$
3
  • $\begingroup$ $0^0$ is also undefined. What is equal to $1$ is the expression $\lim_{x \to 0^+} x^x = 1$. See wolframalpha.com/input/?i=0%5E0 $\endgroup$
    – ArsenBerk
    Commented Jan 23, 2018 at 17:56
  • $\begingroup$ @ArsenBerk Oh I didn't know that. So it may be that these two expressions are always equivalent when defined? $\endgroup$
    – jodag
    Commented Jan 23, 2018 at 18:01
  • $\begingroup$ This is really not about equivalence relations $\endgroup$
    – amrsa
    Commented Jan 23, 2018 at 19:27

1 Answer 1

1
$\begingroup$

Let's try to find it out by considering two cases.

Case 1 ($c \in \mathbb{Z}$): First, let $a = r_1e^{i\theta_1}$ and $b = r_2e^{i\theta_2}$. Then we have $$\bigg(\frac{a}{b}\bigg)^c = \bigg(\frac{r_1e^{i\theta_1}}{r_2e^{i\theta_2}}\bigg)^c = \bigg(\frac{r_1}{r_2}\bigg)^c \cdot e^{i(\theta_1-\theta_2)c} = \frac{r_1^c}{r_2^c}\cdot\frac{e^{i\theta_1c}}{e^{i \theta_2c}} = \frac{a^c}{b^c}$$ so whenever this expression is defined, they are equivalent.

Case 2 ($c \in \mathbb{C}$): We will use the fact:

Let $z,c \in \mathbb{C}$. Then $z^c = e^{c\log(z)}$

We have $$\frac{a^c}{b^c}= \frac{e^{c \log(a)}}{e^{c \log(b)}} = e^{c(\log(a)-\log(b))} = e^{c \log(\frac{a}{b})} = \bigg(\frac{a}{b}\bigg)^c$$ Therefore they are always equivalent when the expression is defined as you suggested.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .