Cambridge 4U questions (1 Viewer)

[richz]

hello,

would anyone help me with chp 2 further questions, qs 8 - 14

thnx

 

[shafqat]

8) gather terms on both sides and take mods, eg: |z1-z2| = |z4 - z3|
that means opposite sides are equal. repeat. remember multiplication by i for the square.

 

[shafqat]

are u happy with help along these lines?

 

[richz]

no dont understand, does ur question read, prove z1 - z2 +z3 - z4 = 0

 

[shafqat]

no dont understand, does ur question read, prove z1 - z2 +z3 - z4 = 0

yes thats the one
z1 - z2 +z3 - z4 = 0
rearranging, z1 - z2 = z4 - z3
taking mods of both sides, |z1 - z2 | = |z4 - z3|
so one pair of opposite sides are equal.
repeating, |z3 - z2| = |z4 - z1|
so the other pair of sides are also equal (remember this defines a parallelogram)
therefore if z1 - z2 +z3 - z4 = 0, then ABCD is a parallelogram

 

[Slidey]

hello,

would anyone help me with chp 2 further questions, qs 8 - 14

thnx

Question 9

z=rcis@
z^2=r^2cis(2@)

Expression becomes:
rcis@/[r^2(cis2@+1)] Go from there.

See the spoiler below for the rest of the working, once you've tried it yourself.

Spoiler

cis@/[r(cis2@+1)]
Multiply top and bottom by [cis(-2@)+1]
[cis(-@)+cis(@)]/[r({cos2@+1}+isin2@)({cos2@+1}-isin2@)]
(2cos@)/[r({cos2@+1}^2+sin^2(2@)]
Which is real.
Simplified further:
2cos@/(r{2+2cos2@})
cos@/(2r.cos^2(@))
1/(2rcos@), the value you're looking for (hopefully)

I think - looks like a bit too much work to me.

EDIT: I hadn't finished putting in the tags, Estel. I have no intention to spoonfeed. Pitty I can't remove the email tags. Gah. Hopefully that will fix it.

 

[Estel]

lol and what precisely is the use of getting the answers spoonfed?

 

[Slidey]

Question 12:
Let z^3=u and expand u^3-1.

 

[Slidey]

13) Textbook exercise. Well, they all are, but this one especially. Both circles, one centre (0,0) and radius 6, the other (5,0) and radius 5.

Range is the possible y values. arg(z) is a ray which starts at origin (origin not included) and extends based on the gradient of z. Use the diagram from part one.

Spoiler

The range of arg(z) based on the two inequalities is the y-values of the points of intersection of both circles. Remember to exclude 0, though, since origin is not a point on the ray.

For the equality part, solve each cricle for y^2, then equate each circle.

 

[Slidey]

14)
a) let z=x+iy. Solve for the pronumeral.
b) see page 68.

 

[Estel]

10,11,12 since nobody's answered

10
simplify by considering 1 + w + w^2 = 0

11
if they have common root, pz^5+qz+r will be zero for one of z = 1, z =w and z=w^2

12
factorise

 

[richz]

thnx guys