anonymous
  • anonymous
how do I solve z^3 =1 using polar representation?
Mathematics
schrodinger
  • schrodinger
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watchmath
  • watchmath
Do you call this as polar representation \(z=re^{i \theta}\) ?
anonymous
  • anonymous
yes!...
anonymous
  • anonymous
\[I have z^3 =1 \to solve letting z=re^ i \Theta\]

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watchmath
  • watchmath
Ok, 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
  • anonymous
\[ \therefore r^3e^(3i \Theta) = 1* e^(2\pi i k) where z is an integer\]
anonymous
  • anonymous
ok, 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
  • watchmath
Remember \(ae^{i\theta}=be^{i\phi}\) 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
  • watchmath
I mean \(r^3e^{3\theta i}=1e^{2\pi ki}\)
anonymous
  • anonymous
\[ok, think I see....OH!!! \] \[e^ (2\pi i k) \]=and z^3 also = 1, THEN they equal each other and ........I get it

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