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anonymous
 5 years ago
how do I solve z^3 =1 using polar representation?
anonymous
 5 years ago
how do I solve z^3 =1 using polar representation?

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watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Do you call this as polar representation \(z=re^{i \theta}\) ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[I have z^3 =1 \to solve letting z=re^ i \Theta\]

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Ok, first recall that \(e^{2\pi ki}=1\) for any k So now we have \(z^3=\left(re^{i\theta}\right)^3=r^3e^{i(3\theta)}=e^{2\pi i}\) Hence \(r=1\) and \(3\theta =2\pi k\) So now \(\theta =(2\pi k)/3\) for \(k=0,1,2\) So we have three solutions: \(1,e^{2\pi i/3},e^{4\pi i/3}\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[ \therefore r^3e^(3i \Theta) = 1* e^(2\pi i k) where z is an integer\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, see you have that also....BUT why is r^3 =1 and e^i(3theta) = e^(2 pi i)? I'm missing something simple here.....

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Remember \(ae^{i\theta}=be^{i\phi}\) iff \(a=b\) and \(\theta=\phi\) Now if we use that for \(r^3e^{i\theta}=1e^{2\pi k i}\) then the conclusion follows :)

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1I mean \(r^3e^{3\theta i}=1e^{2\pi ki}\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[ok, think I see....OH!!! \] \[e^ (2\pi i k) \]=and z^3 also = 1, THEN they equal each other and ........I get it
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