factors of a cubic.
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You got \(\displaystyle \left[\alpha \ = \ 1 \pm \sqrt{3i} \ \right] \)
What did you get for \(\displaystyle \left[ \frac{4}{\alpha} \right] \)?
What did you get for \(\displaystyle \left[ \alpha + \frac{4}{\alpha} + 1\right] \)?
Dr.Peterson
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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].
yeah but there are two values of alpha, are there really 5 roots?
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.
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.
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No. If you have real coefficients, any linear terms involving complex numbers come in conjugate pairs. You can pick either, then the other one will mechanically pop out of the algebra.
It's a long messy problem presented in a deliberately obscure way to get you used to dealing with these concepts.
It's a long messy problem presented in a deliberately obscure way to get you used to dealing with these concepts.
Steven G
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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!
I sympathize. It's why I wait till six to pour a glass.
To keep myself out of trouble during the daytime, I have lately been playing the NANDgame. I already solved it once, but this time I am doing it so as to ensure the fewest possible number of NAND gates. (It was an easy problem for my son, but he is a computer engineer. As a European history major, the game is something of a challenge for me. I don't know what went wrong with my genes: one of my distant cousins named Lehmer was involved in designing computers back in the 1930's and 1940's. )
Dr.Peterson
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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.
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.
This looks so painstaking to type in latex. I must get upto speed with it. Thanks.
Dr.Peterson
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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!).
Great stuff
I shall start swatting these. I was doing long convoluted stuff by googling stuff about latex. I use the [maths][maths] commands.
pka
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Do you understand that \(\dfrac{1}{z}=\dfrac{\overline{~z~}}{|z|^2}~?\)
If so then \(\dfrac{1}{1+\sqrt3\,i}=\dfrac{1-\sqrt3\,i}{4}\).
Harry_the_cat
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conjugate
oh yeah! i remember now
D