Root of Complex Number in Polar Representation with Negative "r"

Question

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.

Answer 1

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 $$ \operatorname{cis}(\tfrac\pi3+\tfrac{2k\pi}3)\operatorname{cis}(\tfrac\pi6+\tfrac{2\ell\pi}3)=\operatorname{cis}(\tfrac\pi3+\tfrac\pi6+\tfrac{2(k+\ell)\pi}3)=\operatorname{cis}(\tfrac\pi2+\tfrac{2(k+\ell)\pi}3) $$

Answer 2

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.