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# Further Maths complex number question watch

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

2. (Original post by RuneFreeze)
Not really sure how to do the first part...

Have you tried anything? Recall that the conjugate of a root is also a root. You can use that to create the q. factor.
3. 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
4. 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.
5. For the result of long division, I got:
$z^2-6z+34+\frac{z(Q+214)}{z^2-4+13}$
It's easy enough to solve for Q from here. I can't see how you could "show that" though.
Edit: -4z, not -4
6. (Original post by Edgemaster)
For the result of long division, I got:
$z^2-6z+34+\frac{z(Q+214)}{z^2-4+13}$
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 is a factor the remainder must be zero.

I.e. the numerator in that last term must be zero.
7. (Original post by ghostwalker)
Since 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
8. wait... Q= negative 214 right?
9. Don't use long division, just compare coefficients,
10. (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.
11. Back! Sorry guys was watching second half of Chelsea Arsenal (gutted)
12. (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.

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?
13. okay finished the question, here's my final answer to anyone interested, although still not really happy the first partAttachment 715226

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