Can someone give me a hint for complex root question

A

annaanna

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The first question is to find k in terms of alpha, 1 is my first attempt and 2 is my second. I tried to write both answers I got from those 2 but it said incorrect. Also I have to cram a lot of maths topics in a short time and I'm finding that the solutions are sometimes methods I knew about but just didn't think to use there because I'm so used to the typical method solutions for each topic,, is there like any way I can get better at remembering techniques other than just doing random practice questions which happen to be trickier. I wish I could just get really good at algebra and come up with the stuff myself but that doesn't seem like something achievable in 3 days 1709632481909.png
 
Just as in Hitchhiker's Guide to the Galaxy, I can see your answers but have no clue what the question is :)

As for better ways for remembering, I am not aware of any formal, fool-proof ways except practice -- sorry :(
 
Either of your approaches should work. In 1):

Can you express [imath]k[/imath] in terms of [imath]a,b,r[/imath] ? Then use the fact that Note that [imath]2arx + a^2x + b^2x = 0[/imath] for every [imath]x[/imath].
Note that you can express both [imath]2a[/imath] and [imath]a^2+b^2[/imath] in terms of [imath]\alpha[/imath] -- do you know how ?
 
blamocur said:
Either of your approaches should work. In 1):

Can you express [imath]k[/imath] in terms of [imath]a,b,r[/imath] ? Then use the fact that Note that [imath]2arx + a^2x + b^2x = 0[/imath] for every [imath]x[/imath].
Note that you can express both [imath]2a[/imath] and [imath]a^2+b^2[/imath] in terms of [imath]\alpha[/imath] -- do you know how ?
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.....If I have to do this in an exam I'll start crying ,I think one of my main problems now is that at this level even if the path i'm taking feels wrong and long it can still be right. Anyways thank you for the hint and guidance !! 1709653870543.png1709653930274.png
 
annaanna said:
.....If I have to do this in an exam I'll start crying ,I
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As long as it does not keep you from solving the problems :)
More seriously, just practice this type of problems some more.
You might find it slightly easier if you express each of the polynomial coefficients through its roots without expanding it first.

You might try finishing your second approach and see if you find it easier. Assuming that [imath]k[/imath] is real (you seem to assume this in 1)) we know that [imath]\gamma = \bar{\beta}[/imath], and thus [imath]k = \alpha + 2\Re \beta[/imath]. You also know that [imath]\alpha|\beta|^2 = 2[/imath] and [imath]\alpha\beta + \alpha\bar\beta + |\beta|^2 = 0[/imath]. Does this look easier to complete?
 
In 2) your multiplying out is not correct. If you try more carefully you should get
x^3-x^2(A+B+C)+x(AB+BC+AC)-ABC

from the question
A+B+C=k (this implies gamma is the complex conjugate of beta)
AB+BC+AC=0 (so AB+Abs(B)^2+AB*=0 so A*2Re(B)+Abs(B)^2=0 )
ABC=2 (so ABB*=2 => A*2Re(B)=2)
 
blamocur said:
As long as it does not keep you from solving the problems :)
More seriously, just practice this type of problems some more.
You might find it slightly easier if you express each of the polynomial coefficients through its roots without expanding it first.

You might try finishing your second approach and see if you find it easier. Assuming that [imath]k[/imath] is real (you seem to assume this in 1)) we know that [imath]\gamma = \bar{\beta}[/imath], and thus [imath]k = \alpha + 2\Re \beta[/imath]. You also know that [imath]\alpha|\beta|^2 = 2[/imath] and [imath]\alpha\beta + \alpha\bar\beta + |\beta|^2 = 0[/imath]. Does this look easier to complete?
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Not at all... (ignore the rectangle that says stuck ) I barely managed to follow through. But I hope in the future I'll at least remember that I can write (whatever the question equivalent for gamma will be) as the conjugate of the other given complex root
1709720778782.png

can i also ask more about the
[imath]k = \alpha + 2\Re \beta[/imath]
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Im assuming since K is real and beta and gamma are conjugates that their v's (from u+vi) cancel out so therefore you can just write it as alpha + 2 real beta? If i noticed this I would've probably written it as alpha + 2u
 
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eipi45 said:
In 2) your multiplying out is not correct. If you try more carefully you should get
x^3-x^2(A+B+C)+x(AB+BC+AC)-ABC

from the question
A+B+C=k (this implies gamma is the complex conjugate of beta)
AB+BC+AC=0 (so AB+Abs(B)^2+AB*=0 so A*2Re(B)+Abs(B)^2=0 )
ABC=2 (so ABB*=2 => A*2Re(B)=2)
Click to expand...
Im a bit confused with the writing but i think youre saying the same thing as person above
 
annaanna said:
Im assuming since K is real and beta and gamma are conjugates that their v's (from u+vi) cancel out so therefore you can just write it as alpha + 2 real beta? If i noticed this I would've probably written it as alpha + 2u
Click to expand...
Correct. I used LaTeX macro \Re which is typeset as [imath]\Re[/imath]. I agree it looks somewhat confusing :(
 
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