In real case, I know that this exponent law holds $x^{ab} =(x^a)^b$.

But if extend the base and exponent to the complex numbers, the above law is not, in general, true.

In text, the author says $z^{ab}$ assumes every value of $(z^a)^b$. and he said that the converse is not true.

I'd like to see how one implication holds but the opposite implication doesn't.

Could you give me analytical argument other than counterexample? (I alreday knew that some counterexample.)

Thank you for your answer in advance.


If $0\ne x\in \Bbb C $ then there there exits $r \in \Bbb R^+$ and $t\in \Bbb R$ such that $x=r\cdot \exp (it+2\pi i n) $ for every $n\in \Bbb Z.$ Even for $0\ne a\in \Bbb R,$ if we let $r^a$ denote its usual value in $\Bbb R ,$ it seems that $x^a$ should be $r^a\cdot \exp (ita+2\pi i n a)$ for $some$ $n\in \Bbb Z.$ But if $a/2 \not \in \Bbb Z$ then there are at least two possible values (and often infinitely many) as $n$ ranges over $\Bbb Z,$ so it is unclear which $n$ is "right". The ambiguity is amplified in $(x^a)^b.$

There is a convention that if $b\in \Bbb N$ then $x^n$ has its intuitive meaning, defined by recursion: $x^1=x$ and $x^{b+1}=x\cdot x^b$ for all $b\in \Bbb N.$ And that if $0\ne x$ and $0>b\in \Bbb Z$ then $x^b=1/(x^{-b}).$ It would be incredibly cumbersome not to have this convention.

It is also common in complex analysis that if $r,a \in \Bbb R$ and $r^a$ is defined in $\Bbb R$ then $r^a$ (in $\Bbb C$) denotes this real value.

This is not a complete explanation.

  • $\begingroup$ Thank you. Then where can I find the "whole" explanation? $\endgroup$ – glimpser May 4 at 18:55
  • $\begingroup$ Generally for $0\ne x\in \Bbb C$ let $S(x)=\{y\in \Bbb C: \exp(y)=x\}.$ For $a\in \Bbb C$ let $T(x,a)=\{\exp (ay):y\in S(x)\}.$ Then $x^a$ is defined as some (any) member of $T(x,a),$ except for the convention in my answer . For example in the (very deep) Gelfond-Schneider Theorem: If $x, a$ are (complex or real) algebraic numbers with $0\ne x\ne 1$ and $a\not \in \Bbb Q$ then (any value of) $x^a$ is transcendental . ( The line "It is also common..." in my A does not apply to this theorem.) $\endgroup$ – DanielWainfleet May 5 at 7:39

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