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z = 4 /(1+ i) Find, in the form a + i b where a, b belong to R, (a) z, (b) z^2. Given that z is a complex root of the quadratic equation x^2 + px + q = 0, where p and q are real integers, (c) find the value of p and the value of q.

a) Need to multiply with conjugate to bring z to form a+ib. => z= z * (1-i)/(1-i) = (4-4i) / 2 = 2-2i b) z^2 = (2-2i)^2 = 4-8i+4 i^2 = 4-8i-4 = 8i since z is root of x^2+px+q=0 then z* (conjugate) is also...
HP
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Show that (n^2) + (n+1)^2 + (n+2)^2 = 3n^2 + 6n + 5, Hence show that the sum of 3 consecutive square numbers is always 2 away from a multiple of 3.

Expanding out the Brackets: (n 2 )+ (n 2 + n + n + 1) + (n 2 + 2n + 2n +4) = (n 2 ) + (n 2 +2n+1) + (n 2 +4n+4) =3n 2 + 6n +5 Using this result: Three consecutice numbers: n, n+1, n+2 So three consecutive sq...
JC
3763 Views

Express cos5x in terms of increasing powers of cosx

De Moivre's theorem: (cos5x + isin5x) = (cosx + isinx) 5 To get cos5x we will need to expand (cosx + isinx) 5 and then take the real parts. Binomial expansion: (cosx + isinx) 5 = (5C0)(cosx) 5 + (5C1)(cosx) ...
CS
36197 Views

Show, using the focus-directrix property for an ellipse, that PS +PS'=2a where P is a point on the ellipse and S and S' are the two foci.

The focus diretrix property for an ellipse is PS/PD=e. Now this is also the case for the other directrix and focus, so PS'/PD'=e. Now we can rearrange these equations to find a formula for PS +PS', PS +PS'=e...
DL
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How do you deal with 3 simultaneous equations? (Struggling with Q7 of AQA specimen paper 1)

If you need to solve them, then you just plug your way through the algebra to get to the answer. In this question (Q7) you need to find the value of a constant such that there is no solution to the three equ...
JW
3786 Views