Further Mathematics: Post your doubts here!

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Hunza01

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does anyone have notes for further mechanics...... please post here...... im in an urgent need of them :(
 
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Hunza01

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does anyone have notes for Further mechanics??
would be very thankful if your can upload it!
 
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Hunza01

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does anyone have notes for further mechanics ...... please post here...... im in an urgent need of them
 
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chris123

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Hi,

Do u have alll the past papers for 2012? I can't find some 2012 on this site.
Also I need further maths (all modules including mechanics) and chemosytry and physics.

Thanks
 
K

Kandinsky

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Help please june 2005 paper 1 question 10(complex)
second part
Click to expand...

You have listed the roots for (a) w^12 = 1.

Now, you have (z+2)^12 = z^12
z = 0 is not a root of the given equation, hence we can divide by z^12, with justification
you get
{(z+2)/z } ^12 = 1
Making a substitution y = (z+2)/z;
you obtain the equation from (a). However, the task is to show that is has 10 non-real roots. Let us find all the roots at first.
y^12 = 1 has 12 roots: 1, -1 and 10 non-real
let us write these roots as (cos A + i*sin A), where A is dependent on natural k (in fact, A = 2pi*k/12, k = 0,6, +-1,2,3,4,5, A is introduced for convenience)
now, after we have found the roots for y, let us find roots z:
(z + 2) / z = cos A + i sin A
then, 1 + 2/z = cos A + i sin A
2/z = cos A + i sin A - 1
z = 2/ (cos A + i sin A - 1)

Multiplying by conjugate, note that conjugate is (cos A - 1 - i sin A), not (cos A + i sin A +1)
[some digression here: we should justify that we are multiplying by a number (cos A - 1 - i sin A). we cannot multiply by 0 (!) so,
cos A - 1 - i sin A = 0 implies that cos A = 1. In turn, A = 0. hence, z = 1, which is NOT a root of the given equation, so cos A - 1 - i sin A is not = 0, hence we justified the multiplication(or we can say that we are not interested in cases when cos A=1, because z = 1 is not a root) ]

z = 2* (cos A - 1 - i sin A) / { (cos A-1)^2 - i^2 * sin^2(A) }
Simplify denominator first,
(cos A - 1)^2 + sin^2(A) = cos(A)^2 + sin^2(A) - 2 cos A + 1 = 2 - 2 cos A = 2 (1 - cos A)
then, z = 2*(cos A - 1 - i sin A) / 2 (1 - cos A)
z = (cos A - 1)/ ( 1 - cos A) - i sin A/ (1 - cos A)

Using formula from trigonometry, tan (A/2) = (1 - cos A)/ sin A; obtain
z = -1 - i*cot(A/2), that is what required. then exclude the possibility A = pi (hence k = 6), since then z = -1 and it is not a non-real number. (so the equation in (b) has 10 non-real roots and -1)
 
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Ashfaq2228

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Does anyone has notes on "using de moivre's theorem in the summation of series"?
please help urgent :s
 
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Heisenberg

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Here's a question for you guys:

A quadratic equation x^2 + bx + c has two real roots which may be distinct or equal. Denoting these roots by r and m, prove that r^(2^n) + m^(2^n) is a polynomial of c in degree 2^(n-1) with leading coefficient 2.
 
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anshul

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hi guys,
i've been working on momentum and impulse,but i'm getting some difficulty concerning coeff. of restitution on inclined planes.
here's an example:
A ball drops vertically onto a smooth plane inclined to the horizontal at an angle α. It hits the plane
with speed 8ms−1 and rebounds horizontally. The coefficient of restitution between the ball and the
plane is 1/3.
Find the value of α and the speed with which the ball rebounds.
(j06/1)
i would really appreciate it if anybody could help me out with this question.
thanks.
 
abcde

abcde

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Q4. Last part which involves 'applying sum of squares to obtain result'. I'm unable to see how 2^2 + 4^2 + ....+ (n-1)^2 = the given result.
 

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