x Turn on thread page Beta
 You are Here: Home >< Maths

# Crazy FP2 questions help watch

1. There's two past paper questions that I simply cannot do, and the mark schemes don't help.

The first question (I can do all but part (c) but included the other parts incase the information in those parts was needed):

(5) The complex number satisfies the relation

(a) Sketch, on an Argand diagram, the locus of . (3 marks) (CAN DO THIS PART)

(b) Show that the greatest value of is . (3 marks) (CAN DO THIS PART)

(c) Find the value of for which

.

It's only part (c) I cannot do... I wrote LOADS of working but the mark scheme does it in a few lines... I wrote z=x+iy therefore (y-4)/(x+4) = tan pi/6 (reversing the arg) therefore y = (x+4) * 1/root3 + 4...

Then I used the modulus formula (mod(z+4-4i) = 4), subbed my z=x+iy into it and reversed that too - squareroot of (x+4)squared + (y-4)squared = 4, then rearranged, subbed my y in from the first part of my working and ended up with x = -4 +/- 2root3 so y = 2, 6.

The correct answer for x was (-4 + 2root 3), and I actually did that, but how come I got (-4 - 2root3) as well? Secondly, my working out took ages where as they did it a strange way by using cos...

Second question: (just part (b) I can't do)

(bored of latex now)

6) It is given that z = e^(itheta).

(a) (i) Show that z + 1/z = 2costheta (2 marks) (CAN DO THIS PART)

(ii) Find a similar expression for z^2 - 1/(z^2) (2 marks) (DONE - 2cos2theta)

(iii) Hence show that z^2 - z + 2 - 1/z + 1/(z^2) = 4cos^2theta - 2 costheta (3 marks) (DONE THIS PART)

(b) Hence solve the quartic equation

z^4 - z^3 + 2z^2 - z + 1 = 0

giving the roots in the form a + ib. (5 marks)

Not a clue. Mark scheme offers 3 solutions, each as crazy as the last...
2. if you multiply by what do you get?
3. (Original post by Pheylan)
if you multiply by what do you get?
How are you meant to know to do that???

Sooo you get (4cos^2theta -2costheta)z^2 = 0

therefore z = 0 or 2costheta(2costheta - 1) = 0

therefore z = 0 or theta = pi/2 or pi/3

so z = 0, e^(ipi/2), e^(ipi/3) ???

The mark scheme says nothing like that... The mark scheme says z = (1 +/- iroot3)/2
4. (Original post by ak9779)
How are you meant to know to do that???

Sooo you get (4cos^2theta -2costheta)z^2 = 0

therefore z = 0 or 2costheta(2costheta - 1) = 0

therefore z = 0 or theta = pi/2 or pi/3

so z = 0, e^(ipi/2), e^(ipi/3) ???

The mark scheme says nothing like that... The mark scheme says z = (1 +/- iroot3)/2
Z=0 is not a solution
Multiplying with z=0 gives always true equation so it is expand the domain of
equation, which may give false solutions.
To check a solution substitute it into the original equation.
It is clear that if z=0 then we get 1=0 which is false.
Similarly for e^(ipi/2).
Note that if costheta=0 then theta=pi/2 or theta=3pi/2
2costheta-1=0 also gives two solutions:
theta=pi/3 and theta=-pi/3 both are true solutions for z.
5. Didn't notice any help for part c of your first question..

Sketch the locus of w when arg(w)=pi/6.

Well then say z+4-4i=w and solve that to get z=w-4+4i. Translate w to get z a sketch of z.

You should have a line from the centre of your circle at pi/6 to the horizontal. A little trigonometry will then give you the answer

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: December 17, 2010
Today on TSR

### Negatives of studying at Oxbridge

What are the downsides?

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams