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Tranformations from z to w plane Watch

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    Need a bit of help with this please.

    From z = x + iy to w = u + iv, transformation T given by w = 1/z. Show that the image, under T, of the line with equation x = 1/2 in the Z plane is a circle C in the w plane. Find equation of C.

    I can't get to first base here. I can do the ones where |z| = 2, or if you have a circle in the Z plane, but I cannot formulate the x = 1/2 - the only thing I can try to do is think about making up the perpendicular bisector to simuklate the locus, such that |z| = |z +1|, but not sure if that helps or not?
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    letting x = 1/2

    you get w = 1/( 1/2 + iy ) now multiply top & bottom by the conjugate of the bottom...

    rearrange to get

    2/(4y2 + 1) - 4iy/(4y2 + 1) **

    now choose some sensible values for y and plot them... you should be able to find the centre and radius of the circle. Then go back and show that the u + vi from ** fit the formula for the circle you have found.
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    Thanks - a little tough - I was hoping I might be able to show something that actually looked like a circle! (at least to me)
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    I hope this helps....
    Attached Images
  1. File Type: pdf Complex plane Transformation Problem.pdf (46.7 KB, 160 views)
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    (Original post by Charries)
    Thanks - a little tough - I was hoping I might be able to show something that actually looked like a circle! (at least to me)
     \displaystyle w = \frac1{z} \implies z = \frac1{w}

    Sub in  w = u + iv , them multiply top and bottom by the conjugate.

    You will get,  \displaystyle z = \frac{u-iv}{u^2 +v^2} = \frac{u}{u^2 + v^2} - \frac{iv}{u^2+v^2}

    We know  z = x + iy = \dfrac12 + iy

     \displaystyle \frac12 + iy = \frac{u}{u^2 + v^2} - \frac{iv}{u^2+v^2}

    Equate the real parts, and you should get the equation of circle.

    Sorry for showing too much working.
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    (Original post by mikelbird)
    I hope this helps....
    Isn't my way better?
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    I would say its the same...eliminating between the variables is the only really complicated part and both of us would have to do it..
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    (Original post by mikelbird)
    I would say its the same...eliminating between the variables is the only really complicated part and both of us would have to do it..
    Next steps in my method are,

     \displaystyle \frac12 + iy = \frac{u}{u^2 + v^2} - \frac{iv}{u^2+v^2}

    Equate real parts,
     \displaystyle \frac12 = \frac{u}{u^2 + v^2} \implies (u-1)^2 + v^2 = 1

    I think that your method is quite complex.
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    (Original post by raheem94)
    Isn't my way better?
    Actually , on second thoughts...it can very much depend on the question....sometimes its easier your way...sometimes mine...!!
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    Thanks to you both. Helps a lot.
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    ... follow up question.

    What if y = 2x + 1 instead - how woud that mapping work?
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    (Original post by Charries)
    ... follow up question.

    What if y = 2x + 1 instead - how woud that mapping work?
     \displaystyle x + iy = \frac{u}{u^2 + v^2} - \frac{iv}{u^2+v^2}

    The above is taken from one of my previous post in this thread.

    Equate the real and imaginary parts,

     \displaystyle x = \frac{u}{u^2 + v^2} \ \ and \ \ y = - \frac{v}{u^2+v^2}

    Insert it in the equation  y = 2x +1

     \displaystyle  - \frac{v}{u^2+v^2}  = 2 \times \frac{u}{u^2 + v^2} + 1

    Now simplify it and you will get an equation of a circle.
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    Raheen94 would give you the answer...
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  2. File Type: pdf Further answer.pdf (35.9 KB, 73 views)
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    Thank you guys - got it.
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    Ok - I have done a few questions like this and everything is going well, but come across this one. w = 16/z, where |z-4|=4.

    Gone through the usual method, converted the thing to a circle in cartesian co-ordinates and get some awful mess of a squared number once I compare real and imaginary parts and substiture for x and y into (what I get) as (u-4)^2 + v^2 = 16.

    Any further clues?
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    z=\frac{16}{w} and |z-4|=4

    \left|\frac{16}{w}-4\right|=4

    \left|\frac{4}{w}-1\right|=1

    \left|\frac{4-w}{w}\right|=1

    |4-w|=|w|

    w=2+vi
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    (Original post by BabyMaths)
    z=\frac{16}{w} and |z-4|=4

    \left|\frac{16}{w}-4\right|=4

    \left|\frac{4}{w}-1\right|=1

    \left|\frac{4-w}{w}\right|=1

    |4-w|=|w|

    w=2+vi
    Nice one...and there is a geometric aspect to this...that last equation before the end is telling you that the locus of the point z is such that it is always equidistant from the point 0 and the point 4 i.e. it is the perpendicular bisector of the two points.

    Anyone like to guess (without calculation) what locus is |z| +|z-4| = 6 ??
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    (Original post by mikelbird)

    Anyone like to guess (without calculation) what locus is |z| +|z-4| = 6 ??
    Thanks.

    Anyone who knows the construction using two pins and a piece of string will know what it is.
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    That's brilliant - thank you.

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    In this mark scheme, have they just got rid of the i because it's a factor of both, and they can pull it out into another modulus and say it's 1. I can see why mod of i squared would be 1 and can be ignored, but I am not convinced why they can do it for i?
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    Actually please explain
 
 
 
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