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    This is Ex. 2C, Question 12 of the Heinemann book (page 43) if anyone's interested. I think I'll just type out the whole question, but please note I only need help on the last part (unless one of the other parts is wrong, of course).

    (i) Show that in an Argand diagram the equation,

    arg(z-2) - arg(z-2i) = 3pi/4, represents an arc of a circle.

    (ii) Show that |z-4|/|z-1| is constant on this circle.

    (iii) Find the values of z corresponding to the points in which this circle is cut by the curve given by,

    |z-1| + |z-4| = 5

    ================================ =================

    (i) Circle with equation x²+y²=4
    I think the arc required is the one such that x+y > 2 (see bolded part of curve in attachment).

    (ii) The expression wittles down to 4, i.e. a constant.

    (iii) Now, I think this is in the form of an equation of an ellipse, with centre (5/2 , 0), and foci at 1 and 4 on the x-axis. The ellipse passes through (0,0) and (5,0) and is at a maximum at (5/2 , ½√91). Is that correct?
    I now wanted to get the equation of the ellipse in its cartesian form (so I could solve), but realised I didn't know how to do it with the centre shifted.
    Any help is appreciated!
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    (ii) should be 2. That is:
    |z-4|/|z-1| = 2
    |z-4| = 2|z-1|

    By substituting this in (iii), you should get:
    3|z-1| = 5
    |z-1| = 5/3

    Which is a circle with center (1, 0) and radius 5/3, i.e. it's equation is:
    (x-1)² + y² = 25/9
    x² - 2x + 1 + y² = 25/9
    4 - 2x + 1 = 25/9 --- (since x²+y²=4)
    x = 10/9

    Use this to solve for y.

    Oh, and I think your sketch is correct.
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    Fantastic! I failed to see the connection between the second and third parts.

    Thanks very much.
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    (Original post by mockel)
    This is Ex. 2C, Question 12 of the Heinemann book (page 43) if anyone's interested. I think I'll just type out the whole question, but please note I only need help on the last part (unless one of the other parts is wrong, of course).

    (i) Show that in an Argand diagram the equation,

    arg(z-2) - arg(z-2i) = 3pi/4, represents an arc of a circle.

    ================================ =================

    (i) Circle with equation x²+y²=4
    I think the arc required is the one such that x+y > 2 (see bolded part of curve in attachment).
    From an algebraic point of view you can do it as follows.

    Remember tan(A-B) = (tanA-tanB)/(1+tanA tanB)

    So

    tanLHS = [y/(x-2) - (y-2)/x]/(1+y(y-2)/{x(x-2)}) = -1 = tanRHS

    yx - (x-2)(y-2) +x(x-2) + y(y-2) = 0

    yx - xy + 2y + 2x - 4 + x^2 + y^2 -2x -2y = 0

    x^2 + y^2 = 4

    which answers the question.

    But if you want to think about it geometrically it's better to think about this equation

    arg(z-2) - arg(z-2i) = 3pi/4

    as a circular arc connecting 2 and 2i which makes a right angle at its centre (from circle theorems) and so the centre is O by inspection
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    Ah, I actually did it the algebraic way, exactly as you've done it, because I was majorly confused as to what it would look like.
    Thanks for the tip for the geometrical way :)
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    arg(z-2) - arg(z-2i) = 3pi/4,

    anyone know what that would look like? been puzzled for a few hours now! and exams on tuesday!!! - i thought it would be an arc on the opposite side - i.e. angle subtended being pi/4 in the 3rd quadrant.

    any clues? - pic would be very appreciated

    phil
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    The pic is attached in mockel's post...
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    (Original post by dvs)
    The pic is attached in mockel's post...
    why is it that - anyone got any logical explanation? - i cant see how that can enclose an angle of 3pi/4 :confused:

    i get it now - how dumb of me, lol - i was confused about what happens when an angle is > pi rad - anyone know what happens tehn, and will we get a question where the arg of two things like that question below = say 3pi/2? etc?
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    It would be the other arc. (See pic - it's the red arc. Btw, this isn't meant to be the exact same pic as mockel's, but I was too lazy to make a new one. The part in black would now represent arg... = pi/2.)

    The reasoning behind this is that we want pi+pi/2, i.e. pi/2 subtended in the 'other' side.
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    (Original post by dvs)
    It would be the other arc. (See pic - it's the red arc. Btw, this isn't meant to be the exact same pic as mockel's, but I was too lazy to make a new one. The part in black would now represent arg... = pi/2.)

    The reasoning behind this is that we want pi+pi/2, i.e. pi/2 subtended in the 'other' side.
    If you sub z =-2 (which is on your arc) into

    arg(z-2) - arg(z-2i)

    you get

    pi - (-3pi/4) = 7pi/4

    which is incorrect (even up to multiples of 2pi)
 
 
 
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