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# Edexcel FP2 Official 2016 Exam Thread - 8th June 2016 watch

1. Does anyone know which paper the de movire's theorem proof was in please? Thanks
2. When asked to find solutions of z and z^5=i, and the range of theta is not stated, can we assume that we have to find all the possible values of z?
3. (Original post by Geraer100)
When asked to find solutions of z and z^5=i, and the range of theta is not stated, can we assume that we have to find all the possible values of z?
Yes I the range is usually between pi and -pi so I would stick with that if no range given
4. (Original post by economicss)
Does anyone know which paper the de movire's theorem proof was in please? Thanks
June 13???? Could be wrong after that
5. (Original post by josephinemar25)
Yes I the range is usually between pi and -pi so I would stick with that if no range given
Thanks!
So you mean that you would find all the possible values?

And when they asked you to shade the region modulus(z)<2, how do you verify/know which part to shade?
6. (Original post by Geraer100)
When asked to find solutions of z and z^5=i, and the range of theta is not stated, can we assume that we have to find all the possible values of z?
I was literally just about to post about this question. The mark scheme lets k = 0, 1, 2, 3, 4. Why not negative values?
7. (Original post by ombtom)
I was literally just about to post about this question. The mark scheme lets k = 0, 1, 2, 3, 4. Why not negative values?
Are you sure there was no range. If not you can either do K=0,1,-1,2,-2 or the positive values, the MS might say oe (or equivalent) and so both would def be accepted.Hope that helps.
8. (Original post by ombtom)
I was literally just about to post about this question. The mark scheme lets k = 0, 1, 2, 3, 4. Why not negative values?
Because the angles eventually begin to repeat themselves so you might as well have a 0 for convenience and avoid negative numbers
For instance, the question 6 of june 2009, why did they shade the area outside?
10. (Original post by josephinemar25)
Are you sure there was no range. If not you can either do K=0,1,-1,2,-2 or the positive values, the MS might say oe (or equivalent) and so both would def be accepted.Hope that helps.
No range; it's the adapted Jan 2006 paper so it might've been cut out by mistake.
11. (Original post by ombtom)
I was literally just about to post about this question. The mark scheme lets k = 0, 1, 2, 3, 4. Why not negative values?
(Original post by Geraer100)
When asked to find solutions of z and z^5=i, and the range of theta is not stated, can we assume that we have to find all the possible values of z?
(Original post by josephinemar25)
Yes I the range is usually between pi and -pi so I would stick with that if no range given
Trigonometric functions are periodic. You can plug in k = -500, -499, -498, -497, -496 for all I care and still get the same answer. They would be accepted unless a specific range was given.
12. Do tricky polar coordinate sketches (eg r^2=a^2 cos(2theta)) ever come up? - I've not seen any in past papers, and haven't learnt them... yet
13. (Original post by Geraer100)

I'm guessing you mean how to shade that on a w plane? Otherwise it's just the inside of the the circle |z|=2.
14. (Original post by economicss)
Does anyone know which paper the de movire's theorem proof was in please? Thanks
(Original post by josephinemar25)
June 13???? Could be wrong after that
Hmm it's not in June 13, there's one on this 2008 paper though...
https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf
I think we're probably due another one this year ☺️
15. I'm a little bit worried since my grades are fluctuating a lot (from Cs to A*s), purely because of silly mistakes :/
16. (Original post by kkboyk)
I'm a little bit worried since my grades are fluctuating a lot (from Cs to A*s), purely because of silly mistakes :/
yeah I feel you, same here. really hoping for a straightforward paper tomorrow.
17. how do you know whether to use x+yi or u+vi
18. (Original post by khaleesi98)
I'm guessing you mean how to shade that on a w plane? Otherwise it's just the inside of the the circle |z|=2.
Yes I mean shading the area on a w plane, its confusing sometimes. What I am doing now is to put the </> instead of equal when working out the cartesian equation of C in w plane and then if the equation of C(assuming a circle) is smaller than 2 ,then shade inside and if bigger, outside. I don't know if this is correct though
19. (Original post by Geraer100)
Yes I mean shading the area on a w plane, its confusing sometimes. What I am doing now is to put the </> instead of equal when working out the cartesian equation of C in w plane and then if the equation of C(assuming a circle) is smaller than 2 ,then shade inside and if bigger, outside. I don't know if this is correct though
Take a point z that satisfies |z| < 2. (say z = 1 + i), plug this into your transformation function. Now, the output, take it and find it's modulus. If |w| < (radius) as well, then you shade inside the circle in the w-plane. If it's the other way around, you shade outside.
20. (Original post by Zacken)
Take a point z that satisfies |z| < 2. (say z = 1 + i), plug this into your transformation function. Now, the output, take it and find it's modulus. If |w| < (radius) as well, then you shade inside the circle in the w-plane. If it's the other way around, you shade outside.
With modulus(w)<radius you mean the radius of the circle in the w plane?

I just have seen a mark scheme and they used what you said but they didn't use the modulus of w, they just choose a z that satisfies modulus(Z)<2 and then take the output w and see if lying inside or outside. Is that the same as using the modulus of w?

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