# Help on Branch Points

1. Jul 16, 2008

### BWV

$(1-z^3)^{1/2}$ has three branch points at $z=1, z=\omega$and $z=\omega^2$ where $\omega = e^{2\pi i/3}$

The branch points are the zeros of the function correct?

So why are not $z=\omega^4$, $z=\omega^6$, $z=\omega^8$ etc also branch points as they are zeros of the function?
as
$(e^{2\pi i/3^3})^6$ =$e^{2\pi i^6}$ = $e^{12\pi i}$ = 1 so all even numbered powers of $\omega$ are zeros of the function

also why does the latex code move around vertically like this?

2. Jul 17, 2008

### tiny-tim

Hi BWV!

$$(\omega^{n})^3 = 1$$ for any n:

$$(\omega^{n})^3 = (e^{\frac{2n\pi}{3}})^3 = e^{2n\pi} = (e^{2\pi})^n = 1$$

If you mean why doesn't it stay on the same line as ordinary text, the answer is:

'cos it doesn't!

But you can make it stay on the line by using [noparse]$and$ instead of $$and$$[/noparse] ("itex" stands for "inline tex") … but that has the disadvantage that it squeezes it vertically into the line, so fractions and powers get squashed.

Alternatively, you can type the whole thing in latex, enclosing each bit of ordinary text inside the brackets of "\text{}"

3. Jul 17, 2008

### HallsofIvy

$\omega= e^{2\pi /3}$ is the "principal cube root" of 1. That is, $\omega^3= 1$. We don't talk about $\omega^4$ because $\omega^4= (\omega^3)(\omega)= (1)(\omega)= \omega$. Similarly, $\omega^6= (\omega)^3(\omega^3)= (1)(1)= 1$ and $\omega^8= (\omega)^3(\omega^3)(\omega^2)= (1)(1)(\omega^2)= \omega^2$.