Turn on thread page Beta
    • Thread Starter
    Offline

    1
    ReputationRep:
    I don't really understand using complex numbers to represent a locus of points on an Argand diagram...


    eg. |z-5-3i|=3
    Does it mean the magnitude of the complex number z minus another complex number (5+3i) equals 3? Or does it mean that the absolute value of the difference of the complex numbers is 3?
    (Or are they the same thing?)

    Furthermore, when you're finding the Cartesian equation of this locus, what exactly are you solving for? And why are we using an x and y axis instead of a real and imaginary one?

    Also :
    |(x-5)+i(y-3)| = 3
    And so the the moduli is removed by squaring :
    (x-5)^2 + (y-3)^2 = 3^2

    What happened to the i?


    thanks
    Offline

    0
    ReputationRep:
    It's a circle, so we need the equation of a circle.

    (x-a)² + (y-b)² = r² where (a,b) is the circle's centre and r is its radius.

    We know that the centre is at (5,3) and the radius is 3. So our equation is (x-5)²+(y-3)²=9
    • PS Helper
    Offline

    14
    PS Helper
    (Original post by confuzzled92)
    eg. |z-5-3i|=3
    Does it mean the magnitude of the complex number z minus another complex number (5+3i) equals 3? Or does it mean that the absolute value of the difference of the complex numbers is 3?
    (Or are they the same thing?)
    The absolute value and modulus of a complex number are the same thing. You can think of complex numbers as vectors, where x+iy corresponds to the vector \begin{pmatrix} x \\ y \end{pmatrix}. With vectors, you should know that, say, \mathbf{a} - \mathbf{b} gives you the vector pointing from the point with position vector \mathbf{b} to the point with position vector \mathbf{a}, and hence |\mathbf{a} - \mathbf{b}| gives you the distance between the two points. It's exactly the same with complex numbers; if w,z are two complex numbers then |w-z| tells you the distance between them on an Argand diagram. So here you have |z-(5+3i)|=3, which tells you that the corresponding locus is the set of all points at distance 3 from the point 5+3i; in other words, it's a circle about 5+3i of radius 3.

    (Original post by confuzzled92)
    Furthermore, when you're finding the Cartesian equation of this locus, what exactly are you solving for? And why are we using an x and y axis instead of a real and imaginary one?
    The Argand diagram (with real and imaginary axes) corresponds to the two-dimensional plane (with x- and y-axes), and so when you find the Cartesian equation of the locus, you stop thinking in terms of z and start thinking in terms of (x,y).

    (Original post by confuzzled92)
    Also :
    |(x-5)+i(y-3)| = 3
    And so the the moduli is removed by squaring :
    (x-5)^2 + (y-3)^2 = 3^2

    What happened to the i?
    You should know the identity |z|^2 = z \bar z, and so if z=a+ib then \bar{z} = a-ib; multiplying the two thus gives a^2+b^2.
    Offline

    8
    ReputationRep:
    (Original post by confuzzled92)
    I don't really understand using complex numbers to represent a locus of points on an Argand diagram...


    eg. |z-5-3i|=3
    Does it mean the magnitude of the complex number z minus another complex number (5+3i) equals 3? Or does it mean that the absolute value of the difference of the complex numbers is 3?
    (Or are they the same thing?)

    Furthermore, when you're finding the Cartesian equation of this locus, what exactly are you solving for? And why are we using an x and y axis instead of a real and imaginary one?

    Also :
    |(x-5)+i(y-3)| = 3
    And so the the moduli is removed by squaring :
    (x-5)^2 + (y-3)^2 = 3^2

    What happened to the i?


    thanks
    You can not remove the moduli by squaring so. THe square of a complex number is another complex number. You would like to get |z|^2 and tried to wrote |z^2|
    |z^2|=|((x-5)+i(y-3))^2|=|(x-5)^2-(y-3)^2+2i(x-5)(y-3)|
    |z^2| \neq |z|^2
    |z|^2=z\cdot \bar{z}=(x-5)^2+(y-3)^2 in your example.
    • PS Helper
    Offline

    14
    PS Helper
    (Original post by ztibor)
    You can not remove the moduli by squaring so. THe square of a complex number is another complex number. You would like to get |z|^2 and tried to wrote |z^2|
    |z^2|=|((x-5)+i(y-3))^2|=|(x-5)^2-(y-3)^2+2i(x-5)(y-3)|
    |z^2| \neq |z|^2
    |z|^2=z\cdot \bar{z}=(x-5)^2+(y-3)^2 in your example.
    As it happens, in fact, |z^2|=|z|^2, but I don't think the OP ever did try to find |z^2| anyway.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: March 29, 2011
Poll
Favourite type of bread
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.