Hello,
I have some troubles understanding why De Moivre's formula for complex roots if valid. It says, if `z = r(cos(\theta) + isin(\theta))` then `z^n = r^n(cos(n\theta) + isin(n\theta))` and we can get roots by: `r^(1/n) (cos((\theta + 2k\pi)/n) + isin((\theta + 2k\pi)/n))`, number of roots -> `k = 0, ..., n-1`.
Now they say "While it is not true that any real number has `n` real nth roots, every complex number (and thus every real number) has n complex roots. " - So what I don't get is why in the same sentence they claim not every real number has nth roots but in the same time, when treated as complex numbers they have?
Thanks
I have some troubles understanding why De Moivre's formula for complex roots if valid. It says, if `z = r(cos(\theta) + isin(\theta))` then `z^n = r^n(cos(n\theta) + isin(n\theta))` and we can get roots by: `r^(1/n) (cos((\theta + 2k\pi)/n) + isin((\theta + 2k\pi)/n))`, number of roots -> `k = 0, ..., n-1`.
Now they say "While it is not true that any real number has `n` real nth roots, every complex number (and thus every real number) has n complex roots. " - So what I don't get is why in the same sentence they claim not every real number has nth roots but in the same time, when treated as complex numbers they have?
Thanks