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I've just stumbled across $$\left(e^{2\pi i}\right)^{\frac{1}{2}}=\sqrt 1=1\neq -1=e^{\pi i},$$do I have some error in my thoughts there or does $(a^b)^c=a^{(bc)}$ not hold for complex numbers?

José Carlos Santos
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RedLantern
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  • No it does not hold. You may find a thread here which explains it. – Wuestenfux Apr 21 '19 at 12:24
  • For example https://math.stackexchange.com/q/2659145/442 or https://math.stackexchange.com/q/1825408/442 or https://math.stackexchange.com/q/1481316/442 – GEdgar Apr 21 '19 at 12:46

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The complex numbers are just a distraction. We can rewrite your example as

$$(-1)^2=1 \implies \big ((-1)^2 \big)^{1/2}=1^{1/2} \implies-1 = \sqrt 1 = 1 \implies -1 =1.$$

Do you see where we made the mistake above?

Daron
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Actually, that equality makes no sense for complex numbers. How do you even define $a^b$ if $a$ and $b$ are arbitrary complex numbers?

José Carlos Santos
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  • Good question. Probably by using some branch of the logarithm that is suited for $a$, say $a^b=\exp(b\text{Log}(a))$. – RedLantern Apr 21 '19 at 12:26
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    But since every non-zero complex number has infinitely many logarithms… – José Carlos Santos Apr 21 '19 at 12:27
  • So what would you say, for example, is $i^i$? It solely depends on the logarithm one wants to use? – RedLantern Apr 21 '19 at 12:28
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    Yes. It takes infinitely many values, depending on the branch of the logarithm you choose. For example, if you choose the principal branch, you get $\operatorname{Log}(i)=\ln 1+\frac{\pi}{2}i=\frac{\pi}{2}i$ and then $i^i=\exp(i\operatorname{Log}(i))=\exp\left(-\frac{\pi}{2}\right)$ (which is real!) – Yuval Gat Apr 21 '19 at 12:33
  • So let's take the principal branch then. Does it make sense to define $a^b=\exp(b\text{Log}(a))$ now? What happens if $a\in\mathbb R^-$? – RedLantern Apr 21 '19 at 12:40
  • Yes, you can define it like that. But then you shouldn't expect to have $(a^b)^c=a^{bc}$ – José Carlos Santos Apr 21 '19 at 13:05
  • Okay, this raises another question though. If I define it like that, I can "prove" $a^{(b+c)}=a^ba^c$. But only for $a$ that are in the slit plane. Can I define $a^b$ in another way so that I can obtain this equailty for other $a$ as well? – RedLantern Apr 21 '19 at 13:09
  • Right now, I don't see how. – José Carlos Santos Apr 21 '19 at 13:15
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That's because $$ (e^{2\pi i})^{1/2}\neq e^{\pi i}, $$ even though their square is equal: $$ (e^{2\pi i})= (e^{\pi i})^2. $$ Just like what's in real numbers, every complex number have two square roots. The power $1/2$ placed at the top right corner just gives you one of them. $e^{\pi i}$ is another square root of $1$.

Ma Joad
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