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

1. watchmath

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

2. anonymous

yes!...

3. anonymous

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

4. 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: $$1,e^{2\pi i/3},e^{4\pi i/3}$$

5. anonymous

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

6. 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.....

7. 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 :)

8. watchmath

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

9. 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