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    Could someone possibly help with 12ii?
    I have no idea where to start


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    (Original post by Mutleybm1996)
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    Could someone possibly help with 12ii?
    I have no idea where to start


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    I need the Cartesian equation found earlier first
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    (Original post by Mutleybm1996)
    Could someone possibly help with 12ii?
    I have no idea where to start
    1. By the way arg works we have: \arg (\frac{z-a}{z-b}) = \theta \Rightarrow \arg(z-a) - \arg(z-b) = \theta

    2. z-a is the complex number that takes us from fixed point a to arbitrary point z. Here z is restricted to the circumference of a circle.

    3. \arg(z-a) measures the angle that this complex number makes with the +ve x-axis.

    4. By circle geometry, you should know that the angle subtended by any point on the circumference of a circle between two fixed points (i.e. in an arc) is constant.

    Does that give you a clue?
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    (Original post by TeeEm)
    I need the Cartesian equation found earlier first
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    here, thank you

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    (Original post by Mutleybm1996)
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    here, thank you

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    arg[(z-a)/(z-b)]=θ

    arg(z-a)- arg(z-b)=θ

    arg(z-6-√3) -arg(z-6+√3) = θ


    sub the point they give you in the above equation to get θ
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    (Original post by TeeEm)
    arg[(z-a)/(z-b)]=θ

    arg(z-a)- arg(z-b)=θ

    arg(z-6-3i) -arg(z-6+3i) = θ


    sub the point they give you in the above equation to get θ
    Thanks
    Could i also ask you about q9?
    I tried some bits and bobs but got stuck

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    (Original post by Mutleybm1996)
    Thanks
    Could i also ask you about q9?
    I tried some bits and bobs but got stuck

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    Posted from TSR Mobile
    my reply is incorrect as I am skipping between pages look at the correction below


    arg[(z-a)/(z-b)]=θ

    arg(z-a)- arg(z-b)=θ

    arg(z-6-3√3) -arg(z-6+3√3) = θ


    sub the point they give you in the above equation to get θ


    you should get theta = pi/2
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    (Original post by TeeEm)
    my reply is incorrect as I am skipping between pages look at the correction below
    Changed it, thank you.
    Could you possibly look at Q9? (In a post earlier)
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    (Original post by Mutleybm1996)
    Changed it, thank you.
    Could you possibly look at Q9? (In a post earlier)
    substitute z = x + iy and carry out the algebra

    this is straight forward


    look at some questions in the link


    http://www.madasmaths.com/archive/ma..._questions.pdf


    there many questions with solutions oncomplex but they are mixed in topics.
    At least 40 questions on complex loci all mixed up/the booklet gets progressively harder.
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    (Original post by TeeEm)
    substitute z = x + iy and carry out the algebra

    this is straight forward


    look at some questions in the link


    http://www.madasmaths.com/archive/ma..._questions.pdf


    there many questions with solutions oncomplex but they are mixed in topics.
    At least 40 questions on complex loci all mixed up/the booklet gets progressively harder.
    The algebra still isn't working out :/


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    (Original post by Mutleybm1996)
    The algebra still isn't working out :/


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    here are the first three lines

    then you should be able to complete it

    |x+iy-k2c|=k|x+iy-c|

    |(x-k2c)+iy|=k|(x-c)+iy|

    (x-k2c)2+y2=k2[(x-c)2+y2]
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    (Original post by TeeEm)
    here are the first three lines

    then you should be able to complete it

    |x+iy-k2c|=k|x+iy-c|

    |(x-k2c)+iy|=k|(x-c)+iy|

    (x-k2c)2+y2=k2[(x-c)2+y2]
    If c=a+bi why have you factored(maybe not the right term) it into the real part of the equation?


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    (Original post by Mutleybm1996)
    If c=a+bi why have you factored(maybe not the right term) it into the real part of the equation?


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    does it say c is complex?
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    (Original post by TeeEm)
    does it say c is complex?
    "Where C=a+bi For any fixed complex number"


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    (Original post by Mutleybm1996)
    "Where C=a+bi For any fixed complex number"


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    sorry I did not see it
    the method is still the same

    put z=x+iy
    c = a +bi

    group real and imaginary inside the mods
    "apply" the mods
    square both sides (careful you do not forget to square the k on the right and then multiply the whole RHS by k2)
    then a big mess to simplify.

    this is nothing but an exercise in concentration.

    good luck
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    (Original post by TeeEm)
    arg[(z-a)/(z-b)]=θ

    arg(z-a)- arg(z-b)=θ

    arg(z-6-√3) -arg(z-6+√3) = θ


    sub the point they give you in the above equation to get θ
    Quick question, why did you substitute those specific values into a and b? Why didn't you switch it round?
    EDIT: sorted that part

    Okay, so I've split it into the two arg sections, what do I do now?

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    (Original post by Mutleybm1996)
    Quick question, why did you substitute those specific values into a and b? Why didn't you switch it round?


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    answer to (a) this is where the arc meets the x axis

    answer to (b) if I put it these values the other way round theta will be -pi/2 which gives exactly the same equation, if you multiply it by -1 (I am not sure if this is what you are asking)
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    (Original post by TeeEm)
    answer to (a) this is where the arc meets the x axis

    answer to (b) if I put it these values the other way round theta will be -pi/2 which gives exactly the same equation, if you multiply it by -1 (I am not sure if this is what you are asking)
    Could you run through what to do after the arg's have been separated? I can't seem to get pi/2


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    (Original post by Mutleybm1996)
    Could you run through what to do after the arg's have been separated? I can't seem to get pi/2


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    wait 2 minutes please
 
 
 
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