According to Wolfram Alpha, $1^z=1$ for $z\in\mathbb{C}$. If this is true, then what is wrong with the following argument that $1^z$ has infinitely many values?

Let $z=x+iy$. Then, \begin{align} 1^z &= 1^{x+iy} \\ &= 1^x \cdot 1^{iy} \\ &= 1^{iy} \\ &= e^{iy\log(1)} \\ \log(1) &= \{0,2i \pi, 4i \pi,6i \pi, \ldots\} \\ 1^z &= \{e^{iy \cdot 0},e^{iy \cdot 2 i \pi},e^{iy \cdot 4 i \pi},e^{iy \cdot 6 i \pi},\ldots\} \\ &= \{e^0,e^{-2y\pi},e^{-4y\pi},e^{-6y\pi},\ldots\} \end{align}

Only one of these values is equal to $1$.

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    $\begingroup$ Wolfram alpha is not God so take its answers in context, not as eternal truths of mathematics; in particular, you are right that $1^z$ is (in general) an infinite set of which $1$ is a distinguished value $\endgroup$ – Conrad Dec 31 '20 at 19:06
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    $\begingroup$ @Jakobian because for some people Wolfram Alpha is God whose answers are eternally true $\endgroup$ – Conrad Dec 31 '20 at 19:07
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    $\begingroup$ @Conrad Thou shall not question the infinite wisdom of WolframAlpha ;) $\endgroup$ – Severin Schraven Dec 31 '20 at 19:09
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    $\begingroup$ @SeverinSchraven I am still waiting for the Beta version... :) $\endgroup$ – Raffaele Dec 31 '20 at 19:10
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    $\begingroup$ If you define $\log$ as a function, we have to pick a single value of $\log 1,$ and we’ll usually choose the value $0.$ If you have $\log$ as a multi valued function, then you are correct, $w^z$ is multi-valued, too. $\endgroup$ – Thomas Andrews Dec 31 '20 at 19:11

If, when $a>0$, you define $a^z$ as $\exp\bigl(z\log(a)\bigr)$, where $\log(a)$ is the only real logarithm of $a$, then indeed we always have $1^z=1$.

Otherwise, it's up to you to tell us how you are defining $1^z$. However, note that the set of all logarithms of $1$ is not $\{0,2\pi i,4\pi,i,6\pi i,\ldots\}$; it's $\{0,\pm2\pi i,\pm4\pi i,\pm6\pi i,\ldots\}$.

  • $\begingroup$ Thank you. I noticed that you once answered a question about a fake proof about why $re^{i\theta}=r$: $$re^{i\theta} = re^{(2 \pi i \theta)/2\pi} = r(e^{2\pi i})^{\theta /2\pi} = r(1)^{\theta/2\pi} = r \, .$$ Is the problem the same as the one here: $e^z$ is a function; but $(e^{2\pi i})^z=1^z$ is not a function if we define $z^w$ as $\exp(w\log(z))$? $\endgroup$ – Joe Dec 31 '20 at 19:23
  • $\begingroup$ No; it has something to do with it, but it's not the same thing. Here, we have a problem of defining the meaning of the expressions that we are using. I provided in my answer the more usual meaning of the expression $a^z$, when $z\in\Bbb C$ and $a\in(0,\infty)$. According to that definition, $1^z$ is indeed $1$, for every $z\in\Bbb C$. If you are using some other definition of $1^z$, then you should tell us which definition you have in mind. $\endgroup$ – José Carlos Santos Dec 31 '20 at 19:29
  • $\begingroup$ I wasn't aware that $a^z$ could have more than one definition if $a$ is a real number. Thank you for clearing up my confusion. In your experience, when people write $e^z$, is this generally just a shorthand for $\exp(z)$? Because if we interpret $e^z$ as $\exp(z\log(e))$, where $\log$ is the complex logarithm, then it would seem that $e^z$ is a multi-valued function. $\endgroup$ – Joe Dec 31 '20 at 19:46
  • $\begingroup$ In my experience, every single time that someone writes $e^z$, what that means is $\exp(z)$. However, concerning $\exp(z\log(e))$, note that I wrote in my answer that I was dealing with the only real logarithm of $a$. In this case, $a=e$, and its only real logarithm is $1$. So, $\exp(z\log(e))=\exp(z)$. $\endgroup$ – José Carlos Santos Dec 31 '20 at 19:49
  • $\begingroup$ Going back to the fake proof: you say that the problem is that in general $e^{ab} \neq (e^a)^b$. In the context of the fake proof, you appear to be saying that $re^{(2 \pi i \theta)/2\pi} \neq r(e^{2\pi i})^{\theta /2\pi}$. Is it possible if you elaborate on what the problem is here? $\endgroup$ – Joe Dec 31 '20 at 20:25

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