Turn on thread page Beta
    • Thread Starter
    Offline

    9
    ReputationRep:
    Can anyone explain to me what I need to do for these three parts and what they're actually asking me to do? I just need some guidance on what I actually need to do in order to solve them.
    Name:  20171023_154649.jpg
Views: 92
Size:  456.1 KB
    • Community Assistant
    Offline

    18
    ReputationRep:
    Community Assistant
    (Original post by MathsMeme)
    Can anyone explain to me what I need to do for these three parts and what they're actually asking me to do? I just need some guidance on what I actually need to do in order to solve them.
    Name:  20171023_154649.jpg
Views: 92
Size:  456.1 KB
    It's pretty clear what you need to do, what bit are you stuck on?
    Offline

    11
    ReputationRep:
    (Original post by MathsMeme)
    Can anyone explain to me what I need to do for these three parts and what they're actually asking me to do? I just need some guidance on what I actually need to do in order to solve them.
    Name:  20171023_154649.jpg
Views: 92
Size:  456.1 KB
    The first two are "locus" problems. The locus of an equation is the set of points which satisfy it i.e. the set of points which lie on the curve defined by it. So you need to be able to figure out which points z in the complex plane the following type of eqn is true:

    |z-a|=|z-b|

    where a and b are fixed complex numbers and z represents any of those complex numbers (i.e. points in the plane) for which that eqn is true.

    Hint: |z-a| is the distance in the plane from a to z.

    Similarly |z-a|=r where r is constant says what about the z's that satisfy it?

    It probably will help you to draw out the complex plane and mark two random points a and b then think about the constraints that the eqns above put on z
    • Thread Starter
    Offline

    9
    ReputationRep:
    (Original post by NotNotBatman)
    It's pretty clear what you need to do, what bit are you stuck on?
    For A, cause it's equal to the modulus of z I'm not sure how it should look sketched out.

    For B, just not sure how I'd go about finding the complex numbers that satisfy both
    • Community Assistant
    Offline

    18
    ReputationRep:
    Community Assistant
    (Original post by MathsMeme)
    For A, cause it's equal to the modulus of z I'm not sure how it should look sketched out.

    For B, just not sure how I'd go about finding the complex numbers that satisfy both
    |z-6| = |z-0|, this is one of the perpendicular bisector forms.

    for part b, it's essentially the points of intersection of the two loci.
    • Thread Starter
    Offline

    9
    ReputationRep:
    (Original post by NotNotBatman)
    |z-6| = |z-0|, this is one of the perpendicular bisector forms.

    for part b, it's essentially the points of intersection of the two loci.
    Would it be the two points of intersection of the circle and the perpendicular bisector? Apologies, I haven't been taught this for some reason
    • Community Assistant
    Offline

    18
    ReputationRep:
    Community Assistant
    (Original post by MathsMeme)
    Would it be the two points of intersection of the circle and the perpendicular bisector? Apologies, I haven't been taught this for some reason
    Yes.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: October 23, 2017

University open days

  1. University of Cambridge
    Christ's College Undergraduate
    Wed, 26 Sep '18
  2. Norwich University of the Arts
    Undergraduate Open Days Undergraduate
    Fri, 28 Sep '18
  3. Edge Hill University
    Faculty of Health and Social Care Undergraduate
    Sat, 29 Sep '18
Poll
Which accompaniment is best?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.