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It is obvious that $\left|e^{v}\right|=e^{\text{Re }v}>0$ showing that $\ln z$ is not defined for $z=0$ .

So the expression $z^{u}=e^{u\ln z}$ cannot be used here.

Nevertheless we don't hesitate to say things like: $0^{3}=0\times0\times0=0$.

Are there some conventions here?

José Carlos Santos
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drhab
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    For integer exponents, when considering fixed exponents, we have unambiguously $0^n = 0$ for $n > 0$, $0^n = \infty$ for $n < 0$, and $0^0 = 1$ by considering the type of the isolated singularity in $0$. For non-integer exponents, one sometimes posits $0^u = 0$ if $\operatorname{Re} u > 0$ and $0^u = \infty$ if $\operatorname{Re} u < 0$. – Daniel Fischer Nov 03 '13 at 19:54
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    I like this convention. $0^b$, for $b$ a positive integer, is $0$. $0^b$, where $b$ is the integer $0$, is $1$. $0^b$ for other $b$ is undefined. And (of course) $a^b$ for nonzero complex $a$ is defined by $\exp(b\log a)$, which may be multi-valued. NOTE: Even if $b$ is the complex number $0$, or the real number $0$, I still want $0^b$ undefined. – GEdgar Nov 03 '13 at 19:58
  • I think it would be absurd not to let $0^3$ equal $0$. Daniel's idea allows this because $\operatorname{Re}(3)>0$. – Stefan Smith Nov 03 '13 at 20:20

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Actually, it is not obvious that $|e^v|=e^{\operatorname{Re}v}$. It is true nonetheless. And the fact that $\ln z$ is not defined when $z=0$ is more basic than that. In fact $\ln z$ is a number such that $e^{\ln z}=z$. But we never have $e^v=0$, since $e^v.e^{-v}=1$.

The concept of $a^n$, when $n\in\mathbb{N}$, is more elementary than the concept of exponential function. It just means$$\overbrace{a\times a\times\cdots\times a}^{n\text{ times}}.$$It turns out that, when $a\neq0$, this is equal to $e^{n\log a}$, where $\log a$ is any logarithm of $a$. But we don't need this to define $a^n$. And, by the definition that I mentioned, it is clear that$$(\forall n\in\mathbb{N}):0^n=0.$$

José Carlos Santos
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  • Thank you. Do you also have an answer to the question in the title? – drhab Nov 01 '17 at 09:28
  • @drhab I think that the only complex numbers $u$ such that $0^u$ is defined are those $u$'s which are real numbers greater than $0$ and that, in that case, $0^u=0$. Thet's because if $u$ is rational and greater than $0$, then you can write $u=\frac pq$, with $p$ and $q$ relatively prime natural numbers, and define, when $x\geqslant 0$, $x^u$ as $\sqrt[q]{x^p}$. With this definition, $0^u=0$ and you can extend the definition to real exponents greater than or equal to $0$ using limits. – José Carlos Santos Nov 01 '17 at 10:07
  • Thank you again. It strikes me that in the comments concerning this question - yours also - I meet the words "I think" or "I like" quite often. Apparantly there is not a great need for conventions on this topic. – drhab Nov 01 '17 at 10:24