How do you know how many solutions there are for z?

[Darkmisc]

Homework Statement:

z^3+8i=0

Relevant Equations:

r^3cis3t = 8cis(-pi/2)

Hi everyone

How do you know how many solutions z has
a) in this problem
b) in general?

I understand that they are rotating 2 pi from (-pi/2) in both directions to get the other two solutions. Should this be done in all problems?
Is it simply a coincidence that there are three solutions when z is in the third power?

If the problem needed to be solved for z^4, would there be four solutions? Or only three (again by rotating 2 pi in both directions)?


Thanks

 

[anuttarasammyak]

As you speculate

"every non-zero, single-variable, degree n polynomial with complex coefficients has, counted with multiplicity, exactly n complex roots. "-- Wikipedia https://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra

If you find n solutions for a z^n equation, you have completed.

 

[fresh_42]

... and since $X^3+a$ and $(X^3+a)'=3X^2$ have no common zeros if $a\neq 0$ we also know that there will be three pairwise distinct roots.