Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    2
    ReputationRep:
    This topic is fairly new to me and I don't know how to go about doing this question:

    Show that the mapping

    w=z+\dfrac{c}{z}

    where  z = x + iy, w=u+iv and  c is a real number, maps the circle |z|=1 in the z plane into an ellipse in the w plane and find its equation.

    I have found u and v in terms of x,y and c but don't know how to progress from here.
    Offline

    0
    ReputationRep:
    Go the other way Find x and y in terms of u and v. Then substitute into |z| =1, i.e. x^2 +y^2 =1 to get a quadratic equation in u and v; hopefully an ellipse.
    Chris
    Offline

    0
    ReputationRep:
    Do you have u=x(1+c)
    and v=y(1-c)?

    Rearrange these to find x and y in terms of u, v and c then use the fact that x^2 + y^2 = 1 Which is the locus of z.
    Offline

    17
    ReputationRep:
    What did you get for u and w? (I would start by assuming |z| = 1 and so writing z = cos t + i sin t).
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by DFranklin)
    What did you get for u and w? (I would start by assuming |z| = 1 and so writing z = cos t + i sin t).
    I got
    u=\dfrac{x}{x+y}(x+y+c) \newline \newline v=\dfrac{y}{x+y}(x+y-c)

    Which looked nasty and then I got scared and came onto TSR

    I though about trying to find u and v in terms of x and y but that seems a little complex (sorry for the awful pun)

    I imagine that the z = cost + i sint is the way to go - I'll give it a whirl and get back to you.
    Offline

    17
    ReputationRep:
    I don't think your functions for u and v are right. Want to post your working?
    Offline

    2
    ReputationRep:
    (Original post by The Muon)
    I got
    u=\dfrac{x}{x+y}(x+y+c) \newline \newline v=\dfrac{y}{x+y}(x+y-c)

    Which looked nasty and then I got scared and came onto TSR

    I though about trying to find u and v in terms of x and y but that seems a little complex (sorry for the awful pun)

    I imagine that the z = cost + i sint is the way to go - I'll give it a whirl and get back to you.
    You should write z = x + iy straight away and use the conjugate to get rid of anything complex in the denominator and you should be able to get w in the form a + bi with a and b some function of x and y. And you can then use the condition abs(z) = 1.

    On an aside the question should have stipulated c is not zero.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by DFranklin)
    I don't think your functions for u and v are right. Want to post your working?
    I'd rather not as it is a huge chunk that looks like it will lead no where but following on from the cost+isint approach.

    I found  u = (1+c)\cos(t) \text{ and } v= (1-c)\sin(t)

    Would this be the equation of the new ellips in parametric form? So what I would have to do is just convert to Cartesian?
    Offline

    0
    ReputationRep:
    (Original post by The Muon)
    I got
    u=\dfrac{x}{x+y}(x+y+c) \newline \newline v=\dfrac{y}{x+y}(x+y-c)

    Which looked nasty and then I got scared and came onto TSR

    I though about trying to find u and v in terms of x and y but that seems a little complex (sorry for the awful pun)

    I imagine that the z = cost + i sint is the way to go - I'll give it a whirl and get back to you.
    I'm pretty sure you should have u=\dfrac{x(x^2+y^2+c)}{x^2+y^2} and v=\dfrac{y(x^2+y^2-c)}{x^2+y^2}
    Since x^2 +y^2 = 1, both expressions can be simplified. From there rearrange for x and y and substitute into x^2 + y^2 = 1

    Spoiler:
    Show
    \dfrac{u^2}{(1+c)^2}+\dfrac{v^2}  {(1-c)^2} = 1 I believe
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Mathletics)
    I'm pretty sure you should have u=\dfrac{x(x^2+y^2+c)}{x^2+y^2} and v=\dfrac{y(x^2+y^2-c)}{x^2+y^2}
    Since x^2 +y^2 = 1, both expressions can be simplified. From there rearrange for x and y and substitute into x^2 + y^2 = 1

    Spoiler:
    Show
    \dfrac{u^2}{(1+c)^2}+\dfrac{v^2}  {(1-c)^2} = 1 I believe
    Yeah, I made a slip in my calculation and forgot to square my x and y when simplifying my denominator.

    I have now got the right answer (albeit by Dave's method)

    Thanks guys - rep will be handed out over the next few days
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    What's your favourite Christmas sweets?
    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
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • 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

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