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first time using this site!

Anyways, my question is why does $i^3$ not equal $(i^2)^{\frac{3}{2}}$? I have just been introduced to complex numbers and the teacher gave us a question to simplify $i^3$ I went and did $i^3=(i^2)*i$ which equals $-i$ but one of my peers used $(i^2)^{\frac{3}{2}}$ in which gives $+i$. Which initially I thought was fine, as I just thought you could have both a negative and positive, but the teacher stated otherwise. He couldn't figure out why, so I tried to prove it with a peer.

Here's where I got to!

RTP: $i^3≠(i^2)^{\frac{3}{2}}$

So $1=\sqrt{1}$ $=\sqrt{-1*-1}$ $=i*i$ $=-1$

Since $1≠-1$, $\sqrt{a}*\sqrt{b}=\sqrt{ab}$, where $a, b \in R$

This then could be expanded to what I initially wanted to prove:

$\sqrt{a}*\sqrt{a}*\sqrt{a}\not= \sqrt{(a)^3}$

Therefore, $\sqrt{-1}*\sqrt{-1}*\sqrt{-1}\not= \sqrt{(-1)^3}$ but can $=i^3$

Therefore, QED.

Upon further thought, I realized this is only true for imaginary numbers and not real numbers, however, something felt off in the first so statement, as it is manipulating the imaginary numbers in a funny way, as you could go between positive 1 and negative 1 'infinitely.' So while I believe that $(i^2)^{\frac{3}{2}}$. I don't quite know why or why my proof is off. So as you do nowadays I threw it into Chatgpt and it came back with imaginary numbers have multifaced values, in which I could kind of of grasp after reading but was hoping for a better explanation here or a different one!!

Thanks in advance if you do :)

EDIT: Thanks everyone for providing links that I can look over! I don't know if I should delete this but in case you stumble upon this and are looking for answers check out: For which complex $a,\,b,\,c$ does $(a^b)^c=a^{bc}$ hold? and Exponent law for complex numbers. Provided by Martin in the comments!

Dominique
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  • For complex numbers the power $z^a$ is only well defined (i.e. has a unique value) when $a$ is an integer. Most notably, rules like $(z_1z_2)^a=z_1^az_2^a$ and $(z^a)^b)=z^{ab)$ won't always hold unless $a$ and $b$ are integers. The theme has been discussed to death for example in the aptly named tag [tag:fake-proofs]. – Jyrki Lahtonen Feb 27 '25 at 07:20
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    Questions essentially equivalent to this have been asked & answered a dozen times on this website. Please try a search. – Gerry Myerson Feb 27 '25 at 07:33
  • Another similar question: https://math.stackexchange.com/q/438/42969 – Martin R Feb 27 '25 at 07:46
  • I don't think this question is a duplicate. It asks about a specific equation that I haven't seen in the standard dupe-magnets. The fact that we have an accidental equality is an additional complication that merits discussion. – Chris Culter Feb 27 '25 at 07:48

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Evaluating whether $i^3\neq (i^2)^{3/2}$ is kind of like evaluating whether $2^3\neq3^2$. In the latter case, you don't need to do a bunch of algebra. It suffices to observe that $2^3=8$, and $3^2=9$, and $8\neq9$. So we conclude that $2^3\neq3^2$ by definition.

In your case, the easy side is $i^3=-i$. The harder side is $z=(i^2)^{3/2}$. Taking the principal branch of the exponent and log functions, we get $z=(-1)^{3/2}=\exp(\frac32\log(-1))=\exp(\frac32\pi i)=-i$.

So, ironically, we have the equality $i^3= (i^2)^{3/2}$ after all! I say this is ironic because the equality is in some sense accidental. There was no a priori reason to believe it's true. In fact, you can check that $i^3\neq (i^4)^{3/4}$.

To summarize, your peer made two mistakes:

  1. Introducing $(i^2)^{3/2}$ without an argument that its value is relevant
  2. Evaluating $(i^2)^{3/2}$ incorrectly

In some sense, these two mistakes "cancelled out", and the end result was to illustrate the true fact that complex exponentiation is tricky. It's just the details that were wrong.

Chris Culter
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