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    Hi all, I'm doing this question:

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    I can do part a and b fine, but not c.

    I made z the subject, then plugged in z=x+iy and w =u+iv then rationalised the denominator and equated the imaginary part to 0. but it doesn't seem to give me the right circle equation. I end up with a weird circle equation in terms of u and v.

    Am I working backwards or something?
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    (Original post by Sir_Malc)
    Hi all, I'm doing this question:

    Name:  Maths.jpg
Views: 86
Size:  47.8 KB

    I can do part a and b fine, but not c.

    I made z the subject, then plugged in z=x+iy and w =u+iv then rationalised the denominator and equated the imaginary part to 0. but it doesn't seem to give me the right circle equation. I end up with a weird circle equation in terms of u and v.

    Am I working backwards or something?
    There was no need to introduce w = u+iv, just z = x+ iy. This should clear it all up!
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    (Original post by Sir_Malc)
    Hi all, I'm doing this question:

    Name:  Maths.jpg
Views: 86
Size:  47.8 KB

    I can do part a and b fine, but not c.

    I made z the subject, then plugged in z=x+iy and w =u+iv then rationalised the denominator and equated the imaginary part to 0. but it doesn't seem to give me the right circle equation. I end up with a weird circle equation in terms of u and v.

    Am I working backwards or something?
    Messy way: set z = a + i b and find out what the imaginary part of \frac{z+i}{3+iz} is in terms of a and b. Now do the same for |z-i| = 2.
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    (Original post by Zacken)
    There was no need to introduce w = u+iv, just z = x+ iy. This should clear it all up!
    How come you only need the z=x+iy?

    In the Edexcel book there is similar examples and they always seem to rearrange the transformation first to make z the subject. then plug in z and w.
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    (Original post by Gregorius)
    Messy way: set z = a + i b and find out what the imaginary part of \frac{z+i}{3+iz} is in terms of a and b. Now do the same for |z-i| = 2.
    Not so messy because you're already asked to find the cartesian equation in part (a).

    Out of interest, is there a nice way to do stuff like this? I've glanced over a few complex analysis notes and they always talk about "circles going to lines" and "lines going to lines" or "mobious transforms" or such and I thought it might be vaguely related to this sort of thing where circles in the z-plane go to lines in the w-plane, etc...
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    (Original post by Sir_Malc)
    How come you only need the z=x+iy?

    In the Edexcel book there is similar examples and they always seem to rearrange the transformation first to make z the subject. then plug in z and w.
    Well, I suppose you could plug in u,v but it would serve no purpose since you'll end up only considering v=0 anyway. The video I linked works through this precise example, if you're interested.
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    (Original post by Zacken)
    Not so messy because you're already asked to find the cartesian equation in part (a).

    Out of interest, is there a nice way to do stuff like this? I've glanced over a few complex analysis notes and they always talk about "circles going to lines" and "lines going to lines" or "mobious transforms" or such and I thought it might be vaguely related to this sort of thing where circles in the z-plane go to lines in the w-plane, etc...
    Yes, mobius transformations do have very nice properties, one of which is to map (generalized) circles to (generalized) circles - where a generalized circle is either a common or garden circle or a circle of infinite radius (i.e. a straight line). Once you know this, the question can be dispatched into the long grass by inspecting the fate of three points.
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    (Original post by Gregorius)
    Yes, mobius transformations do have very nice properties, one of which is to map (generalized) circles to (generalized) circles - where a generalized circle is either a common or garden circle or a circle of infinite radius (i.e. a straight line). Once you know this, the question can be dispatched into the long grass by inspecting the fate of three points.
    Ahhh. Yes, I think I've heard about that - you usually try and pick 'easy points' like 0 or infinity or 1 (where infinity is the thingy at the top of the Riemann sphere thingymabob) - it's all very interesting, cheers!
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    (Original post by Zacken)
    Well, I suppose you could plug in u,v but it would serve no purpose since you'll end up only considering v=0 anyway. The video I linked works through this precise example, if you're interested.
    So I should get the same answer even if I rearrange the transformation to z = (i-3w)/(wi-1)

    then plug in z=x+iy and w=u+iv

    and then equate the imaginary part to 0. Then I should get a similar equation but in terms of u and v right?
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    (Original post by Sir_Malc)
    So I should get the same answer even if I rearrange the transformation to z = (i-3w)/(wi-1)

    then plug in z=x+iy and w=u+iv

    and then equate the imaginary part to 0. Then I should get a similar equation but in terms of u and v right?
    You'll get it in terms of u and v, but that's going to serve of no use to you because you want to prove a fact about the (x-y) plane and not the (u-v) plane.
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    (Original post by Zacken)
    You'll get it in terms of u and v, but that's going to serve of no use to you because you want to prove a fact about the (x-y) plane and not the (u-v) plane.
    Thankyou. This makes sense to me now/
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    (Original post by Sir_Malc)
    Thankyou. This makes sense to me now/
    Cheers!
 
 
 
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