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I'm studying Complex Analysis, and I've seen the definition of the set-valued power function as follows

Let $z,w \in \mathbb{C}$, then $z^{w} \equiv \exp(w\log z)$.

If I recall correctly. Now it seems there is something wrong with this definition, because you can't use it to define powers of $0$, which should naturally be $0$. Am I missing something? Or is the definition 'weird'?

user3813284
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Note that powers like $0^{-1}$ or $0^0$ are not defined. The formula $z^w=e^{w\log z}$ is only valid for $z\neq 0$.

ajotatxe
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    $0^{-1}$ is not defined, but $0^0=1$ by several definitions. – hmakholm left over Monica Mar 15 '15 at 01:06
  • Shouldn't we expect for $x \in \mathbb{R}$ and $w \in \mathbb{N}$ that the definition of $x^{w}$ agrees with the usual definition for exponentiation for the reals? If we take the principal value at least. EDIT: Or are mathematicians really don't mind not defining powers of $0$? – user3813284 Mar 15 '15 at 01:12
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    @user3813284: The usual situation is that if we have two different definitions of $a^b$ that both give a value for a particular $a$ and $b$, then we expect these to agree. But there's no single nice definition that gives meaning to all the cases we want to use. (See here for a short survey of some possible definitions of $a^b$). – hmakholm left over Monica Mar 15 '15 at 01:16
  • @HenningMakholm So what you're saying is that, for instance, using the usual definition gives meaning to complex numbers raised to rational powers and $0^{p/q}$, whereas this definition allows us to raise to complex numbers, and we have to sort of choose and compromise? – user3813284 Mar 15 '15 at 01:22
  • @user3813284: What I'm saying is that there's no single "the usual definition". There are several different defintions with different domains. For an input where more than one of them give a value, the value happens to agree. Therefore it is usual (and unproblematic) to use the same notation for all of them. – hmakholm left over Monica Mar 15 '15 at 01:26