De Movire's Theorem: Applications for Root Extraction

  • Context: Undergrad 
  • Thread starter Thread starter revitgaur
  • Start date Start date
  • Tags Tags
    Theorem
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 1K views
revitgaur
Messages
3
Reaction score
1
I want to know about applications of De Movire's theorem for root extraction.
 
Mathematics news on Phys.org
Perhaps the greatest value of De Moivre's Theorem, is the ability to find the n distinct roots of a complex number.

Let ##z = p(cos\theta + isin\theta)## and let ##z^{n} = w##. Then if ##w = r(cos\phi+ isin\phi)##, ##z^{n} = [p(cos\theta + isin\theta)]^{n}## we have
that ##p^{n}[cos(n\theta) +isin(n\theta)] = r(cos\phi+ isin\phi)##. That implies that ##p^{n} = r## and ##n\theta = \phi##, or equivalently ##p =
\sqrt[\leftroot{-2}\uproot{2}n]{r}## and ##\theta = \frac{\phi}{n}##. But ##sin## and ##cos## have period of ##2\pi## so ##n\theta = \phi + 2\pi k## or equivalently ##\theta = \frac{\phi + 2\pi k}{n}##, ##k = 0, 1, 2, \cdots, n - 1##. If we set ##k = n## the solutions are repeated . So, for a positive integer ##n##, we find ##n## distinct ##n-th## roots for the complex number ##w = r(cos\phi+ isin\phi)## : ##z = \sqrt[\leftroot{-2}\uproot{2}n]{r} [cos \frac{\phi + 2\pi k}{n} + isin \frac{\phi + 2\pi k}{n}]##.

With the help of the above formula, we can compute the ##n- th## roots of e.g ##1## or maybe some specific roots, e.g ##4 - th##.
 
Last edited:
  • Like
Likes   Reactions: revitgaur
QuantumQuest said:
Perhaps the greatest value of De Moivre's Theorem is the ability to find the n distinct roots of a complex number.

Let ##z = p(cos\theta + isin\theta)## and let ##z^{n} = w##. Then if ##w = r(cos\phi+ isin\phi)##, ##z^{n} = [p(cos\theta + isin\theta)]^{n}## we have
that ##p^{n}[cos(n\theta) +isin(n\theta)] = r(cos\phi+ isin\phi)##. That implies that ##p^{n} = r## and ##n\theta = \phi##, or equivalently ##p =
\sqrt[\leftroot{-2}\uproot{2}n]{r}## and ##\theta = \frac{\phi}{n}##. But ##sin## and ##cos## have period of ##2\pi## so ##n\theta = \phi + 2\pi k## or equivalently ##\theta = \frac{\phi + 2\pi k}{n}##, ##k = 0, 1, 2, \cdots, n - 1##. If we set ##k = n## the solutions are repeated . So, for a positive integer ##n##, we find ##n## distinct ##n-th## roots for the complex number ##w = r(cos\phi+ isin\phi)## : ##z = \sqrt[\leftroot{-2}\uproot{2}n]{r} [cos \frac{\phi + 2\pi k}{n} + isin \frac{\phi + 2\pi k}{n}]##.

With the help of the above formula we can compute the ##n- th## roots of e,g ##1## or maybe some specific roots e.g ##4 - th##.
Thanks for helping me
 
  • Like
Likes   Reactions: QuantumQuest