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How does one plot $z=(-2)^{3/5}$?

I see in Wolfram Alpha:

https://www.wolframalpha.com/input/?i=z%5E5%3D-8

But I don't understand how that plot is created.

mavavilj
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  • What do you know of complex numbers? In particular, do you know how to plot them in the complex plane? Do you know of the polar form $z=re^{i\theta}$? – Ross Millikan May 28 '19 at 14:41
  • @RossMillikan I know how to plot a complex number $x+iy$. I don't know why I see numerous points for complex number that has only real part. – mavavilj May 28 '19 at 14:42

1 Answers1

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In the complex numbers a number has five fifth roots. If we use the polar form, we have $z=re^{i \theta}$ and we want $z^{5}=-8$ so $$\left(re^{i \theta}\right)^{5}=-8\\ r^{5}=8\\r=8^{1/5}=2^{3/5}\\ \left(e^{i\theta}\right)^{5}=-1\\ 5\theta=\pi+2k\pi\\ \theta=\frac \pi 5,\frac {3\pi}5, \frac {5\pi}5, \frac {7\pi}5,\frac {9\pi}5$$ The complex solutions are shown on the page. You can get approximate values by clicking the box.

Ross Millikan
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  • How you know if have five roots? – mavavilj May 28 '19 at 16:09
  • $k$ is some integer. It reflects the fact that rotation of a multiple of $2\pi$ puts you back where you were. In the complex plane an $n^{th}$ degree polynomial always has $n$ roots if you include multiplicity. Here we have demonstrated them by finding all the values that can be generated by different $k$. The $n^{th}$ roots of a complex number always fall on a circle centered on the origin and are equally spaced on it, as here. – Ross Millikan May 28 '19 at 16:12
  • Why is $5 \theta = \pi + 2k \pi$? – mavavilj May 28 '19 at 16:17
  • Because the angle of $-1$ is $\pi$ – Ross Millikan May 28 '19 at 16:30
  • I don't understand. I understand that angle of $-1$ is $\pi$, but not where $5 \theta=...$ comes from. – mavavilj May 28 '19 at 16:33
  • $5\theta$ comes from the previous line, where $(e^{i\theta})^5=e^{5i\theta}$. The exponential of a pure imaginary is on the unit circle at the angle of the imaginary. Remember that $e^{i \pi}=-1$, which is what says $-1$ is at angle $\pi$ – Ross Millikan May 28 '19 at 18:25