cant find zero values

[hazy fern]

For some reason I can’t find the zero values for these equations

[lunar tartan]

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[cerulean orchid]

(1): try the integer root theorem.
(2): recall how to find the roots of unity.

[hazy fern]

I’ll try

[molten cliff]

A general tip for this kind of question is that you can look for obvious solutions

[molten cliff]

you did quite well here

[hazy fern]

The problem is that I don’t find imaginary values eventhough the solution says so

[molten cliff]

well

it's a bit hard to read, but your intuition about factorizing $(\lambda + 1) P_2(\lambda)$ is correct

[weak heath]

Rion

[molten cliff]

So all that remains is that you find out $P_2(\lambda)$ and factorize it if possible

[weak heath]

Rion

[molten cliff]

I see that you think $P_2(\lambda) = -4 \lambda^2 + 8\lambda$ ? That seems quite unlikely

[weak heath]

Rion

[molten cliff]

because then 0 would be a root, which is clearly not the case

[hazy fern]

I have to find lambda = 2+-2i

[molten cliff]

Yeah, then you have to find the correct $P_2(\lambda)$

[weak heath]

Rion

[hazy fern]

Oh wait

I messed up the symbols

4lamba2 - 8lamba

And I messed up the rule🫠

[molten cliff]

Just for accuracy, suppose that $P_2(\lambda) = a\lambda^2 + b \lambda + c$, then $(\lambda +1)P_2(\lambda) = a \lambda^3 + (b+a) \lambda^2 + (c+b) \lambda + c$

[weak heath]

Rion

[molten cliff]

so evidently, a = 1, b+a = -3, c+b = 4, c = 8

yup, that looks a lot better

$P_2(\lambda) = \lambda^2 - 4 \lambda + 8$

[weak heath]

Rion

[hazy fern]

Yeah I see what I got wrong

[molten cliff]

Now you can compute the complex roots of that

and you should find exactly what they intended you to find

[hazy fern]

How do I start with the second question?

Lamba3 - 1 = 0

[molten cliff]

As Darpinger suggested, this is just a question about the cubic roots of unity

[cerulean orchid]

Well, that is just λ^3 = 1.
Do you know how to solve z^n = w, where n is a positive integer and w is a complex number?

[molten cliff]

Otherwise you can also proceed with a factorization

[cerulean orchid]

Ah, yeah, true. That also works in this case.

[hazy fern]

I don’t think so

[cerulean orchid]

Ah, that's fine. Then factor λ^3 - 1 first.

[hazy fern]

How

Do u mean like this?

[molten cliff]

No

You see that 1 is an obvious solution

[hazy fern]

Yeah

[molten cliff]

So again, you can factor it as $(\lambda - 1) P_2(\lambda)$

[weak heath]

Rion

[molten cliff]

What is $P_2(\lambda)$ ?

[weak heath]

Rion

[hazy fern]

(Lamba - 1)Lamba

[molten cliff]

No

[hazy fern]

No wait

[molten cliff]

it's a quadratic

which leading coefficient is obvously 1

[cerulean orchid]

I recommend recalling how to factorize x^n - 1.

[molten cliff]

and constnat coeff is also clearly 1

[molten cliff]

also incorrect

[hazy fern]

Yeah +

Instead of minus

[molten cliff]

$P_2(\lambda) = \lambda^2 + \lambda + 1$

[weak heath]

Rion

[molten cliff]

in general, $x^n - 1 = (x-1)\sum_{k=0}^{n-1} x^k$

Please learn that by heart

[hazy fern]

👍

Ty

[molten cliff]

So now you can find the complex roots of $P_2$

[weak heath]

Rion

[hazy fern]

Yes

Ty

[weak heath]

Rion

[molten cliff]

Sorry, dumb mistake

I forgot but the sum stops at n-1

As a proof:
$$(x-1) \sum_{k=0}^{n-1} x^k = \sum_{k=0}^{n-1} (x^{k+1} - x^k)$$

[weak heath]

Rion

[molten cliff]

Which is a telescoping sum

[cerulean orchid]

Or we can also look at the sum, which is a geometric progression.

[molten cliff]

unfortunately not working when $x = 1$

[weak heath]

Rion

[cerulean orchid]

Well, works in the limit. x^n - 1 is clearly divisible by x - 1, anyway.

[hazy fern]

Ty

+close