# Thread: Cubic equation (complex numbers)

1. ## Cubic equation (complex numbers)

Hi everyone, this is my first post here. If this belongs in a different section, my apologies.

Question: P(z) := z^3 + pz + q, p and q are elements of Real numbers, u and v and Ϛ are elements of the complex numbers.

u^3 = (-q/2) + sqrt((q/2)^2 + (p/3)^3) and v^3 = (-q/2) - sqrt((q/2)^2 + (p/3)^3)

and Ϛ^3 = 1

Show that z1 = u + v, z2 =
Ϛu + (Ϛ^2)v and z3 = (Ϛ^2)u + Ϛv fufill the condition that P(z1) = P(z2) = P(z3) = 0

I have the following:

u^3 + v^3 = -q

and (u*v)^3 = (u^3)*(v^3)
= ((-q/2) + sqrt((q/2)^2 + (p/3)^3))*((-q/2) - sqrt((q/2)^2 + (p/3)^3))
= ((-q^2)/4)) - ((p^3)/27)

and

P(z1) = (u+v)^3 + p(u+v) + q
= u^3 + v^3 + 3((u^2)*v + u*(v^2)) + p(u+v) + q
= 3((u^2)*v + u*(v^2)) + p(u+v)

I don't know how to continue.

Thanks for any help.

2. Originally Posted by Olym
Hi everyone, this is my first post here. If this belongs in a different section, my apologies.

Question: P(z) := z^3 + pz + q, p and q are elements of Real numbers, u and v and Ϛ are elements of the complex numbers.

u^3 = (-q/2) + sqrt((q/2)^2 + (p/3)^3) and v^3 = (-q/2) - sqrt((q/2)^2 + (p/3)^3)

and Ϛ^3 = 1

Show that z1 = u + v, z2 = Ϛu + (Ϛ^2)v and z3 = (Ϛ^2)u + Ϛv fufill the condition that P(z1) = P(z2) = P(z3) = 0

I have the following:

u^3 + v^3 = -q

and (u*v)^3 = (u^3)*(v^3)
= ((-q/2) + sqrt((q/2)^2 + (p/3)^3))*((-q/2) - sqrt((q/2)^2 + (p/3)^3))
= ((-q^2)/4)) - ((p^3)/27)

and

P(z1) = (u+v)^3 + p(u+v) + q
= u^3 + v^3 + 3((u^2)*v + u*(v^2)) + p(u+v) + q
= 3((u^2)*v + u*(v^2)) + p(u+v)

I don't know how to continue.

Thanks for any help.
Let's do it by parts: first, it's clear u^3 + v^3 = -q, so:

P(z1) (u + v)^3 + p(u + v) + q = u^3 + v^3 + 3uv(u + v) + p(u + v) + q

= (p + 3uv)(u + v).

Now just check the first factor above p + 3uv is zero!

With z2 and z3 is simmilar.

Tonio

3. Yes, this looks like an attempt to present Cardan's Formula.

If you want to see a full exposition of his method of solving cubics, you can find it here:

Cardan's Formula - ProofWiki

4. Originally Posted by Matt Westwood
Yes, this looks like an attempt to present Cardan's Formula.

Don't make it french, he was italian: it is Gerolamo Cardano

Tonio

If you want to see a full exposition of his method of solving cubics, you can find it here:

Cardan's Formula - ProofWiki
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5. Yes all right so he was Italian, but when I did the initial research, this formula was reported on Wikipedia as "Cardan's formula". I suppose it must have been updated since. (It's immaterial anyway, he wasn't actually the one who discovered it.)

6. Originally Posted by tonio
Let's do it by parts: first, it's clear u^3 + v^3 = -q, so:

P(z1) (u + v)^3 + p(u + v) + q = u^3 + v^3 + 3uv(u + v) + p(u + v) + q

= (p + 3uv)(u + v).

Now just check the first factor above p + 3uv is zero!

With z2 and z3 is simmilar.

Tonio
OK, unfortunately I don't understand what you mean by: "Now just check the first factor above p + 3uv is zero!".

Another question (related to this one): If Ϛ^3 = 1 (and it's a complex number), then Ϛ^2 = 1 and Ϛ = 1, right?

Thanks

7. Originally Posted by Olym
Another question (related to this one): If Ϛ^3 = 1 (and it's a complex number), then Ϛ^2 = 1 and Ϛ = 1, right?

Thanks
Not necessarily. Consider $\displaystyle \frac {-1 + i \sqrt 3}{2}$. And of course its conjugate $\displaystyle \frac {-1 - i \sqrt 3}{2}$.

8. Originally Posted by Olym
OK, unfortunately I don't understand what you mean by: "Now just check the first factor above p + 3uv is zero!".

This means: prove that p + 3uv is zero...what else?

Tonio

Another question (related to this one): If Ϛ^3 = 1 (and it's a complex number), then Ϛ^2 = 1 and Ϛ = 1, right?

Thanks
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9. Originally Posted by Matt Westwood
Not necessarily. Consider $\displaystyle \frac {-1 + i \sqrt 3}{2}$. And of course it's conjugate $\displaystyle \frac {-1 - i \sqrt 3}{2}$.
D'oh

Originally Posted by tonio
.........................................
Sorry about that, I just misread what you wrote. It's been a long day...