when w^3 = 1 is it true in each complex number and what exactly is standard form and omega here?

[full karma]

complex number doubt

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[simple swallow]

...what?

[livid canopy]

when w^3 = 1 is it true in each complex number and what exactly is standard form and omega here?
hey OP wtf does this mean

[full karma]

i mean how w^3 = 1 is valid in those complex numbers which are not cube root of unity like here

[livid canopy]

that is a cube root of unity

multiply it by z+1

[simple swallow]

That's still not an explanation.

[dreamy mason]

because it is just negated version

think of it like this

if we have $z_1^3=1$, and we have the equation $z^3=-1$ (this comes from the post you made earlier), what do you think we can do to either of these equations to make it match up

[livid canopy]

wait no it's not sorry that's a 6th root of unity

[shut oar]

ooaa

[livid canopy]

z^3 is like -1 here

[dreamy mason]

yes

[dreamy mason]

$\omega^3=1$ usually represents a cube root of unity

[shut oar]

ooaa

[dreamy mason]

standard form of $\omega$ is $e^{i\frac{\pi}{3}k}=\cos(\frac{\pi k}{3})+i\sin(\frac{\pi k}{3})$, $k\in\bZ$

[shut oar]

ooaa

[dreamy mason]

this also comes from $z^3+1=(z+1)(z^2-z+1)$

[shut oar]

ooaa

[dreamy mason]

i hope this clears your doubts

[full karma]

here if i put k = 3 it becomes e^i = -1 where is w here ?

[dreamy mason]

e^(i*pi) rather

$\omega=e^{\frac{i\pi k}{3}}$, i was just showing standard form

[shut oar]

ooaa

[dreamy mason]

from this def

[dreamy mason]

however if $z^3=-1$ then $-z^3=1$ and $(-z)^3=1$, if we let $w=-z$ we have $w^3=1$, and get the cube roots of unity as desired

[shut oar]

ooaa

[full karma]

so is it also cube root of unity?

@dreamy mason

[dreamy mason]

rather, a negated version

as per the subsitution

[full karma]

gotcha thanks

+close

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