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FP1 Complex Numbers Question

Hi, I am stuck on this problem, because I don't really know how to start...
If 3-2i and 1+i are two of the roots of the equation ax4 + bx3 +cx2 + dx + e = 0, what are the values of a, b c, d and e?

Any suggestions would be greatly appreciated!
Complex roots occur in conjugate pairs; that is, if x=p+qi is a root, then x=p-qi is also a root.

You can figure out all four roots like that, and write the equation as a product of its factors; if x=p+qi is a root, then (x-(p+qi)) is a factor. Then just multiply out the brackets to find the coefficients.

Hope that helps, quote me if it needs more explaining!
Original post by Katie21v9
Hi, I am stuck on this problem, because I don't really know how to start...
If 3-2i and 1+i are two of the roots of the equation ax4 + bx3 +cx2 + dx + e = 0, what are the values of a, b c, d and e?

Any suggestions would be greatly appreciated!


This polynomial isn't uniquely defined. There are infinitely many possible values for the coefficients, even if a is given.

Original post by Mostly_Crazy
Complex roots occur in conjugate pairs


This can only be used if the coefficients of the polynomial are real.
Original post by morgan8002
This polynomial isn't uniquely defined. There are infinitely many possible values for the coefficients, even if a is given.



This can only be used if the coefficients of the polynomial are real.


I am aware of both, but do bear in mind this is an FP1 question. The method is described is perfectly adequate, assuming that what the OP wrote is the text of the question in full.
Reply 4
Original post by Mostly_Crazy
Complex roots occur in conjugate pairs; that is, if x=p+qi is a root, then x=p-qi is also a root.

You can figure out all four roots like that, and write the equation as a product of its factors; if x=p+qi is a root, then (x-(p+qi)) is a factor. Then just multiply out the brackets to find the coefficients.

Hope that helps, quote me if it needs more explaining!


Thanks :smile: So are the four roots:
3-2i, 3+2i, 1+i, 1-i ?
Then do I basically use the factor theorem?
Original post by Katie21v9
Thanks :smile: So are the four roots:
3-2i, 3+2i, 1+i, 1-i ?
Then do I basically use the factor theorem?


Yes; as @morgan8002 clarified, if we know/assume the polynomial has only real coefficients, those are the roots.

So you can write the polynomial as (x - (3+2i)) (x - (3-2i)) (x - (1+i)) (x - (1-i))

Then just multiply out all the brackets to find the coefficients (or rather, one possible set of coefficients - which is most likely all that's necessary at this level).
pfeeew
I thought there was going to be a fight.
Original post by spurs_boi1999
pfeeew
I thought there was going to be a fight.


:biggrin:

No need for a fight, everything he said was right, and things like that SHOULD be pointed out to students sooner rather than later - though unfortunately, they're often not. Which, thinking about it, is probably an example of "teaching to the test"; that sort of thing isn't necessary to pass an FP1 exam, so teachers don't bother with it when teaching FP1 :s-smilie:
Reply 8
Original post by Mostly_Crazy
Yes; as @morgan8002 clarified, if we know/assume the polynomial has only real coefficients, those are the roots.

So you can write the polynomial as (x - (3+2i)) (x - (3-2i)) (x - (1+i)) (x - (1-i))

Then just multiply out all the brackets to find the coefficients (or rather, one possible set of coefficients - which is most likely all that's necessary at this level).


Thankyou :smile: I got to this point, but then I was unsure, as it seems to limit the value of a to 1
Original post by Katie21v9
Thankyou :smile: I got to this point, but then I was unsure, as it seems to limit the value of a to 1


This is related to what Morgan was saying - there are actually an infinite number of different sets of coefficients, because is you multiply every term by the same amount the roots stay the same. Think about it as a stretch transformation; the curve gets stretched up and down, but the points where it intercepts the x-axis stay the same. So if you (for example) replaced the first bracket with (2x - 2(3+2i)), you'd get a different quartic equation with the same roots, but each coefficient would be multiplied by 2.

So yes, writing it in the form I gave does indeed limit a to 1, but for an FP1 question, that will be what they're expecting you to do, if nothing else is stated in the question.
Reply 10
Original post by Katie21v9
Thankyou :smile: I got to this point, but then I was unsure, as it seems to limit the value of a to 1


Original post by Mostly_Crazy

So yes, writing it in the form I gave does indeed limit a to 1, but for an FP1 question, that will be what they're expecting you to do, if nothing else is stated in the question.


Not sure why you're using "limited" if f(x)=0f(x) = 0 where ff is a quartic polynomial, then for all real λ0\lambda \neq 0, we have λf(x)=0\lambda f(x) = 0 and if you decided to give the values of the coefficients as λa1,λb1,\lambda a_1, \lambda b_1, \cdots the markscheme would definitely accept that.
Original post by Zacken
Not sure why you're using "limited" if f(x)=0f(x) = 0 where ff is a quartic polynomial, then for all real λ0\lambda \neq 0, we have λf(x)=0\lambda f(x) = 0 and if you decided to give the values of the coefficients as λa1,λb1,\lambda a_1, \lambda b_1, \cdots the markscheme would definitely accept that.

True, but let's be honest, who actually does that in an exam when just going with a=1 will get you full marks?
Reply 12
Original post by Mostly_Crazy
True, but let's be honest, who actually does that in an exam when just going with a=1 will get you full marks?


Guilty, especially when some of the coefficients are rational, then I prefer multiplying through by the denominator to make the whole thing integral.
Original post by Zacken
Guilty, especially when some of the coefficients are rational, then I prefer multiplying through by the denominator to make the whole thing integral.


*shrug*

Fair enough
Reply 14
Original post by Mostly_Crazy
:biggrin:

No need for a fight, everything he said was right,


i missed the fun again ...:smile:
Reply 15
I feel like using
Unparseable latex formula:

\[x^{2} - (\alpha +\beta )x + \alpha \beta\]

where alpha and beta are the roots to the solution is a quicker way to find the quadratic.
(edited 8 years ago)

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