You got \(\displaystyle \left[\alpha \ = \ 1 \pm \sqrt{3i} \ \right] \)I have worked these out but I cant seem to understand the final ans. How did the come up with 3? I tried using 4/[MATH]\alpha[/MATH] and substituted the roots but no luck.View attachment 23963View attachment 23964
yeah but there are two values of alpha, are there really 5 roots?They presumably used the expressions [MATH]\frac{4}{\alpha}[/MATH] and [MATH]\alpha+\frac{4}{\alpha}+1[/MATH] to find the other roots after finding [MATH]\alpha[/MATH].
OK, I get the point. For the record I do know that but after some JD & coke I did not see it. Damn you got me good!It's a long messy problem presented in a deliberately obscure way to get you used to dealing with these concepts.
I sympathize. It's why I wait till six to pour a glass.OK, I get the point. For the record I do know that but after some JD & coke I did not see it. Damn you got me good!
Did you try actually doing it before asking??? Rather than ask someone else, find out for yourself.yeah but there are two values of alpha, are there really 5 roots?
Okay, but keep going! Don't stop after one step. At least check your work.[MATH]4/\alpha[/MATH] gives [MATH]-1/2 -3/2 i[/MATH] using [MATH]\alpha[/MATH] is [MATH]1+\sqrt 3 i[/MATH]
Sorry I did. I have made mistakes so the roots were not as per the question. I need to show workings but as always,i try to save time by just posting the Q with no workings. Many thanks. I think that 1/2 -3/2i is clearly wrong. I should sleep and try again in the morning.Did you try actually doing it before asking??? Rather than ask someone else, find out for yourself.
Okay, but keep going! Don't stop after one step. At least check your work.
An essential part of learning mathematics well is to try before giving up. You very often discover that what happens is not what you think (that's called learning!). In this case, each choice of alpha leads to the other choice coming from 4/alpha, and both then lead to the third value being 3.
And, of course, surprises are fun. So be bold and explore the new territory.
This looks so painstaking to type in latex. I must get upto speed with it. Thanks.You need to know the Fundamental Theorem of Algebra. If a polynomial is of degree n, it can be factored into a constant times n linear terms, some, none, or all of which which may contain complex numbers. Furthermore, if one of those linear terms is x - u, then u is a root (a zero) of that function.
You have a cubic. That is a polynomial of degree three. Thus, it can be factored into three linear terms.
[MATH]f(x) = ax^3 + bx^2 + cx + d \text { and } a \ne 0 \implies \exists \ u,\ v,\ w \in \mathbb C \text { such that }\\ f(u) = 0,\ f(v) = 0,\ f(w) = 0 \text { and } f(x) = a(x - u)(x - v)(x - w).[/MATH]That is the general principle. Got it?
You are given the specific example:
[MATH]f(z) = z^3 + pz^2 +qz - 12 = 1(x - \alpha)\left(x - \dfrac{4}{\alpha} \right ) \left \{ x - \left ( \alpha + \dfrac{4}{\alpha} + 1 \right ) \right \}\ =[/MATH]
[MATH]x^3 - x^2 * \text{stuff} - x * \text {other stuff} - \alpha * \dfrac{4}{\alpha} * \left ( \alpha + \dfrac{4}{\alpha} + 1 \right ).[/MATH]
But the last term in that mess is just a constant so it must be true that
[MATH]- \alpha * \dfrac{4}{\alpha} * \left ( \alpha + \dfrac{4}{\alpha} + 1 \right ) = - 12.[/MATH]
[MATH]\therefore 4 * \left ( \alpha + \dfrac{4}{\alpha} + 1 \right ) = 12 \implies[/MATH]
[MATH]\alpha + \dfrac{4}{\alpha} + 1= 3 \implies \alpha^2 + 4 + \alpha = 3 \alpha \implies.[/MATH]
[MATH]\alpha^2 - 2 \alpha + 4 = 0 \implies \alpha = \dfrac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(4)}}{2 * 1} \implies[/MATH]
[MATH]\alpha = \dfrac{2 \pm \sqrt{4 - 16}}{2} = \dfrac{2 \pm \sqrt{- 12}}{2} \implies[/MATH]
[MATH]\alpha = \dfrac{2 \pm \sqrt{4 * (- 1)(3)}}{2} = \dfrac{2 \pm 2i\sqrt{3}}{2} = 1 \pm i\sqrt{3}.[/MATH]
Let's arbitrarily decide to pick [MATH]\alpha = 1 + i\sqrt{3}.[/MATH]
[MATH]\therefore \dfrac{4}{\alpha} = \dfrac{4}{1 + i\sqrt{3}} = \dfrac{4}{1 + i\sqrt{3}} * \dfrac{1 - i\sqrt{3}}{1 - i\sqrt{3}} \implies[/MATH]
[MATH]\dfrac{4}{\alpha} = \dfrac{4(1 - i\sqrt{3}}{1 - i^2 * 3} = \dfrac{4(1 - i\sqrt{3}}{1 - (-3)} = 1 - i\sqrt{3}.[/MATH]
[MATH]\therefore \alpha + \dfrac{4}{\alpha} + 1 = 1 + i \sqrt{3} + 1 - i\sqrt{3} + 1 = 1 + 1 + 1 = 3.[/MATH]
Now go back and figure out what stuff and other stuff are in terms of alpha, and you will know what p and q equate to.
This looks so painstaking to type in latex. I must get upto speed with it. Thanks.
`f(z) = z^3 + pz^2 +qz - 12 = 1(x - \alpha)(x - 4/\alpha ) { x - (\alpha + 4/\alpha + 1 ) } =`
`x^3 - x^2 * \text{stuff} - x * \text {other stuff} - \alpha * 4/\alpha * ( \alpha + 4/\alpha + 1 ).`
...
Let's arbitrarily decide to pick `\alpha = 1 + i sqrt{3}.`
`\therefore 4/\alpha = 4/(1 + i sqrt(3)) = 4/(1 + i\sqrt(3)) * (1 - i sqrt(3)/(1 - i sqrt(3))) \implies`
`4/\alpha = (4(1 - i sqrt(3))/(1 - i^2 * 3)) = (4(1 - i sqrt(3))/(1 - (-3))) = 1 - i sqrt(3).`
`\therefore \alpha + 4/\alpha + 1 = 1 + i sqrt(3) + 1 - i sqrt(3) + 1 = 1 + 1 + 1 = 3.`
AsciiMath is a lot easier to learn (though it still requires careful proofreading!).[MATH]f(z) = z^3 + pz^2 +qz - 12 = 1(x - \alpha)\left(x - \dfrac{4}{\alpha} \right ) \left \{ x - \left ( \alpha + \dfrac{4}{\alpha} + 1 \right ) \right \}\ =[/MATH]
[MATH]x^3 - x^2 * \text{stuff} - x * \text {other stuff} - \alpha * \dfrac{4}{\alpha} * \left ( \alpha + \dfrac{4}{\alpha} + 1 \right ).[/MATH]...
Let's arbitrarily decide to pick [MATH]\alpha = 1 + i\sqrt{3}.[/MATH]
[MATH]\therefore \dfrac{4}{\alpha} = \dfrac{4}{1 + i\sqrt{3}} = \dfrac{4}{1 + i sqrt{3}} * \dfrac{1 - i\sqrt{3}}{1 - i\sqrt{3}} \implies[/MATH]
[MATH]\dfrac{4}{\alpha} = \dfrac{4(1 - i\sqrt{3}}{1 - i^2 * 3} = \dfrac{4(1 - i\sqrt{3}}{1 - (-3)} = 1 - i\sqrt{3}.[/MATH]
[MATH]\therefore \alpha + \dfrac{4}{\alpha} + 1 = 1 + i \sqrt{3} + 1 - i\sqrt{3} + 1 = 1 + 1 + 1 = 3.[/MATH]
I'd like to use this occasion to demonstrate a simplified form of LaTeX that you may find less intimidating, called AsciiMath, which I only recently realized is supported here.
Here is a chunk of what JeffM wrote, rewritten in AsciiMath:
`f(z) = z^3 + pz^2 +qz - 12 = 1(x - \alpha)(x - 4/\alpha ) { x - (\alpha + 4/\alpha + 1 ) } =``x^3 - x^2 * \text{stuff} - x * \text {other stuff} - \alpha * 4/\alpha * ( \alpha + 4/\alpha + 1 ).`...Let's arbitrarily decide to pick `\alpha = 1 + i sqrt{3}.``\therefore 4/\alpha = 4/(1 + i sqrt(3)) = 4/(1 + i sqrt(3)) * (1 - i sqrt(3))/(1 - i sqrt(3)) \implies``4/\alpha = (4(1 - i sqrt(3))/(1 - i^2 * 3)) = (4(1 - i sqrt(3))/(1 - (-3))) = 1 - i sqrt(3).``\therefore \alpha + 4/\alpha + 1 = 1 + i sqrt(3) + 1 - i sqrt(3) + 1 = 1 + 1 + 1 = 3.`
This is all I had to type to say that, using backslashes to mark some special commands and characters, and putting backquotes around it:
Code:`f(z) = z^3 + pz^2 +qz - 12 = 1(x - \alpha)(x - 4/\alpha ) { x - (\alpha + 4/\alpha + 1 ) } =` `x^3 - x^2 * \text{stuff} - x * \text {other stuff} - \alpha * 4/\alpha * ( \alpha + 4/\alpha + 1 ).` ... Let's arbitrarily decide to pick `\alpha = 1 + i sqrt{3}.` `\therefore 4/\alpha = 4/(1 + i sqrt(3)) = 4/(1 + i\sqrt(3)) * (1 - i sqrt(3)/(1 - i sqrt(3))) \implies` `4/\alpha = (4(1 - i sqrt(3))/(1 - i^2 * 3)) = (4(1 - i sqrt(3))/(1 - (-3))) = 1 - i sqrt(3).` `\therefore \alpha + 4/\alpha + 1 = 1 + i sqrt(3) + 1 - i sqrt(3) + 1 = 1 + 1 + 1 = 3.`
For comparison, here is the original LaTeX output:
AsciiMath is a lot easier to learn (though it still requires careful proofreading!).
Do you understand that \(\dfrac{1}{z}=\dfrac{\overline{~z~}}{|z|^2}~?\)[MATH]4/\alpha[/MATH] gives [MATH]-1/2 -3/2 i[/MATH] using [MATH]\alpha[/MATH] is [MATH]1+\sqrt 3 i[/MATH]
what does that line mean on top of z. I wasnt aware.Do you understand that \(\dfrac{1}{z}=\dfrac{\overline{~z~}}{|z|^2}~?\)
If so then \(\dfrac{1}{1+\sqrt3\,i}=\dfrac{1-\sqrt3\,i}{4}\).
oh yeah! i remember nowconjugate
.... Watch it! Watch that profanity! Children are listening......... Damn you got me good!