A friend had been looking at, as an example, $z^3=-8cis(\frac{\pi}{2})$ and ran into a phenomenon he struggled with explaining to himself; he approached me for assistance and I wasn't sure, either.

If I look at the "-" as a $cis(\pi)$ and perform a complex multiplication, I end up with $z = 8(cis(\pi)\cdot cis(\frac{\pi}{2})) = 8cis(\frac{3\pi}{2}) \Rightarrow z = 2cis(\frac{\pi}{2}+\frac{2\pi}{3}\cdot k)$ with k being 0,1,2. Wolfram tells me this is the correct answer, and the whole process feels quite logical and intuitive to me.

But we tried another thing: to take the root without converting the minus to a $cis(\pi)$. Basically, looking at it like a "negative module". The actual idea was: if I take the third root of the whole expression, then I can split it into the third root of minus one times the third root of $8cis(\frac{\pi}{2})$ which should give me the same result (or so we thought), but this leads me to $-2cis(\frac{\pi}{6}+\frac{2\pi}{3}\cdot k)$, and even if I do perform the minus-to-cis conversion now, I get a different result.

I've been trying to do two things:

  1. Explain to myself intuitively why this is wrong, or what even is the meaning of the alternative answer that I get.
  2. Write & examine the proof I know for the formula of a complex number's nth root and attempt to algebraically explain to myself why a negative "r" might invalidate it (or: why I must first convert it into the form of module (which has to be positive) times cis(angle), and strictly that form, before taking the root).

Neither of those was I very successful with doing. I'd really appreciate assistance.

  • $\begingroup$ I've generally been quite strict, when opting to work with numbers in their polar representation, with making sure I perform the conversion into a module * cis(angle) form before doing anything else, because that way I felt like I could account for everything; and it's been working out just fine. But... now that I realized things actually do get wonky if I don't do this, I really want to know why. :) Thanks in advance! (& Sorry for any incorrect terminology; not a native English speaker.) $\endgroup$ – ShyGuy Jun 27 '20 at 22:02

It's true that one cannot assume that roots are multiplicative over the complex numbers. But I don't think that's an issue here.

You don't show your second computation. I have $$ -\operatorname{cis}(\tfrac\pi2)=\operatorname{cis}(\pi)\,\operatorname{cis}(\tfrac\pi2). $$ If you write the cubic roots the usual way, you get $$ \operatorname{cis}(\tfrac\pi3+\tfrac{2k\pi}3)\operatorname{cis}(\tfrac\pi6+\tfrac{2\ell\pi}3)=\operatorname{cis}(\tfrac\pi3+\tfrac\pi6+\tfrac{2(k+\ell)\pi}3)=\operatorname{cis}(\tfrac\pi2+\tfrac{2(k+\ell)\pi}3) $$

  • $\begingroup$ Oh, wow. I feel a bit silly now. I'm gonna blame it on the the fact I wrote this at 2 AM. :) Thanks a lot! That said, I'm glad I wrote it because your answer makes it easier for me to understand exactly what confused me. Just to clarify: does the lack of multiplicativity stem from the fact that there's more than just one root for every number, and only specific combinations of the roots are actually the root of the product? (i.e, in your example, the actual roots are those given specifically if l=k.) $\endgroup$ – ShyGuy Jun 28 '20 at 8:56
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    $\begingroup$ Yes. The prototypical example of root bad behaviour is $$1=\sqrt1=\sqrt{(-1)(-1)}``="\sqrt{-1}\sqrt{-1}=-1.$$ $\endgroup$ – Martin Argerami Jun 28 '20 at 14:55
  • $\begingroup$ Awesome. Thanks a lot! $\endgroup$ – ShyGuy Jul 1 '20 at 15:55

The two solutions seem to actually be the same, but with different values of $k$ for the same roots. For example, cis$(\frac\pi6+\frac{2\pi}32)=-$cis$(\frac\pi2)$. I suggest that you plot the first solution, labeling each of those three points with the corresponding value of $k$. Then do the same for the second solution. I think you'll immediately see what's going on and why both solutions are correct.

  • $\begingroup$ Thank you very much :) $\endgroup$ – ShyGuy Jun 28 '20 at 8:48

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