Did I prove this right? (1 Viewer)

[calamebe]

Hey, so basically I'm really not sure if my method for question 2. i) is right at all. I'm confident with the other parts, but if I've failed somewhere just let me know. Anyway album to working and question is here: http://m.imgur.com/a/1lSPf

Thanks in advance!

 

[calamebe]

Might as well lump this in this thread, for 3. (ii) in this question, shouldn't it be 2(p^1/2)cos((2m(pi)±Φ)/3). What happened to the minus? http://m.imgur.com/NOyrzrv

 

[InteGrand]

Hey, so basically I'm really not sure if my method for question 2. i) is right at all. I'm confident with the other parts, but if I've failed somewhere just let me know. Anyway album to working and question is here: http://m.imgur.com/a/1lSPf

Thanks in advance!

You seemed to have gone the other way. The Q. basically wants us to assume those conditions on s and t, and show that x is a solution if this is the case.

If s and t satisfy those conditions, then your line before you 'equated coefficients' is clearly a true statement, and then reversing your steps shows that s+t is a solution of the given cubic equation.

 

[eyeseeyou]

What topic is this?

 

[InteGrand]

Might as well lump this in this thread, for 3. (ii) in this question, shouldn't it be 2(p^1/2)cos((2m(pi)±Φ)/3). What happened to the minus? http://m.imgur.com/NOyrzrv

Don't need minus, since m is just an integer, so it can be negative too, taking care of your worry about the minus.

 

[InteGrand]

What topic is this?

4U Polynomials probably.

 

[calamebe]

Nah, it's some random complex analysis textbook haha, hence why it's here. Anyway thanks heaps for your help.

 

[calamebe]

Also for this one: http://m.imgur.com/jVMRqRd I have an infinite series for the real term and the imaginary term, but don't know how to work it out without wolfram alpha.

 

[calamebe]

Oh wait nevermind I found out how to do it.