The Student Room Group

Further Maths complex number question

Not really sure how to do the first part...

screenshot.1.jpg
(edited 6 years ago)
Original post by RuneFreeze
Not really sure how to do the first part...

screenshot.1.jpg


Have you tried anything? Recall that the conjugate of a root is also a root. You can use that to create the q. factor.
Reply 2
yeah, thank you. I know this I already found the other quadratic factor like you would with any other question like this. I then tried dividing the function by this new quadratic z^2 - 4z + 13 but ran into problems. I found the factor given in the question but couldn't finish the long division because of the Q, so couldn't prove a 0 remainder.

Basically I'm confused about the "show that" part
Update: I was wrong but I'm getting somewhere

Update: Ok! So as z=2-3i was a root, z=2+3i (the conjugate) as Edgemaster says, is also a root. Therefore, (z-2+3i) and (z-2-3i) must be factors. Multiply the brackets out and you get (z^2 -4z +13) is a factor. Next step I guess is algebraic long division!

Update: Alright, so algebraic long division worked! So proud of myself right now, ahaha (I'm not doing further maths) so I proved z^2 -6z+34 was a factor. I'm guessing you know how to do algebraic long division, so basically put z^2 -4z +13 on the 'outside' and f(z) on the inside. Do the division, and on the top you should get the factor! And doing the division makes it easy to see how to do part b.

Update: So I got Q= 214? Hopefully you will get the same!

Update: (Last one I promise!) I found the other roots of the equation to be z=3+5i and z=3-5i

... Sorry I didn't see your last post @RuneFreeze: you don't need to prove the remainder is zero, as long as you have 442 as the constant I think it's OK. You just need to show that the factor will be on the top. It only asks you to find Q in part b, so I think you can get away with not having zero as the remainder in part a.
(edited 6 years ago)
For the result of long division, I got:

It's easy enough to solve for Q from here. I can't see how you could "show that" though.
Edit: -4z, not -4
(edited 6 years ago)
Original post by Edgemaster
For the result of long division, I got:

It's easy enough to solve for Q from here. I can't see how you could "show that" though.
Edit: -4z, not -4


Since z24z+13z^2-4z+13 is a factor the remainder must be zero.

I.e. the numerator in that last term must be zero.
Original post by ghostwalker
Since z24z+13z^2-4z+13 is a factor the remainder must be zero.

I.e. the numerator in that last term must be zero.


Yeah, but I was talking about part a), ignore me I was just being dumb
wait... Q= negative 214 right?
Don't use long division, just compare coefficients,
Reply 9
Original post by Sanjith Hegde123
Don't use long division, just compare coefficients,

Yeah, when asking you to prove that something is a factor, comparing coefficients is quicker and easier.
Back! Sorry guys was watching second half of Chelsea Arsenal (gutted)
I'll read the responses now
Original post by Keen Bean #1
Update: I was wrong but I'm getting somewhere

Update: Ok! So as z=2-3i was a root, z=2+3i (the conjugate) as Edgemaster says, is also a root. Therefore, (z-2+3i) and (z-2-3i) must be factors. Multiply the brackets out and you get (z^2 -4z +13) is a factor. Next step I guess is algebraic long division!

Update: Alright, so algebraic long division worked! So proud of myself right now, ahaha (I'm not doing further maths) so I proved z^2 -6z+34 was a factor. I'm guessing you know how to do algebraic long division, so basically put z^2 -4z +13 on the 'outside' and f(z) on the inside. Do the division, and on the top you should get the factor! And doing the division makes it easy to see how to do part b.

Update: So I got Q= 214? Hopefully you will get the same!

Update: (Last one I promise!) I found the other roots of the equation to be z=3+5i and z=3-5i

... Sorry I didn't see your last post @RuneFreeze: you don't need to prove the remainder is zero, as long as you have 442 as the constant I think it's OK. You just need to show that the factor will be on the top. It only asks you to find Q in part b, so I think you can get away with not having zero as the remainder in part a.


Thanks for your response

I don't see why you don't have to prove the remainder is zero because otherwise you can't be sure it's a factor, no?
okay finished the question, here's my final answer to anyone interested, although still not really happy the first part
Attachment not found
(edited 6 years ago)

Quick Reply

Latest