Question

how do I solve z^3 =1 using polar representation?

Answer 1

Do you call this as polar representation \(z=re^{i \theta}\) ?

Answer 2

yes!...

Answer 3

\[I have z^3 =1 \to solve letting z=re^ i \Theta\]

Answer 4

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: \(1,e^{2\pi i/3},e^{4\pi i/3}\)

Answer 5

\[ \therefore r^3e^(3i \Theta) = 1* e^(2\pi i k) where z is an integer\]

Answer 6

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.....

Answer 7

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 :)

Answer 8

I mean \(r^3e^{3\theta i}=1e^{2\pi ki}\)

Answer 9

\[ok, think I see....OH!!! \] \[e^ (2\pi i k) \]=and z^3 also = 1, THEN they equal each other and ........I get it