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    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.
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    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
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    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.
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    What did you get for u and w? (I would start by assuming |z| = 1 and so writing z = cos t + i sin t).
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    (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.
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    I don't think your functions for u and v are right. Want to post your working?
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    (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.
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    (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?
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    (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:
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    \dfrac{u^2}{(1+c)^2}+\dfrac{v^2}  {(1-c)^2} = 1 I believe
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    (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
 
 
 
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