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    Hey Guys

    I'm a bit stuck on this question :confused:

    Identify the locus in the complex plane given by |z + i| = 2

    What does 'identify' mean. I've got the equation of a circle, but it includes 'i'. Can this be correct?

    Express the complex number z = -i + 1/(1 - i), stating the values of x and y. Find the modulus and argument of z and plot  \overline{z} on an Argand diagram.
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    Identify means find.
    It might be correct, but only if the 'i's cancel! In the Cartesian plane, equations of circles are always real. What is your equation?

    ...What is the problem with the last one? Multiply top and bottom of the fraction by the complex conjugate of what's on the bottom. Rearrange. Done. Do you know how to put it into polar form?
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    x^2 + y^2 + 2xi - 5 = 0

    D'oh! The other question seems so obvious now. Thanks amigo
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    (Original post by The Silk Cobra)
    x^2 + y^2 + 2xi - 5 = 0

    D'oh! The other question seems so obvious now. Thanks amigo
    Hmm... Well that doesn't look quite right.

    What is the centre of the circle in the question? And the radius? Put that into the equation for a circle.
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    radius = 2

    centre = (-i,0)
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    (Original post by The Silk Cobra)
    radius = 2

    centre = (-i,0)
    Is that for the first question because that's wrong.
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    (Original post by StephenP91)
    Is that for the first question because that's wrong.
    Yes
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    (Original post by The Silk Cobra)
    Yes
    |Z + i| = 2

    Let  Z = x + iy

    So the modulus of Z + i is \sqrt{x^{2} + (y + 1)^{2}}

    So \sqrt{x^{2} + (y + 1)^{2}} = 2 \rightarrow x^{2} + (y + 1)^{2} = 4

    So circle centre (0,-1), radius 2.
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    In this situation, x= Re(z) and y=Im(z). So since the centre is at -i, this corresponds to (0,-1). Can you see this?
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    (Original post by meatball893)
    In this situation, x= Re(z) and y=Im(z). So since the centre is at -i, this corresponds to (0,-1). Can you see this?
    Yeah, that's what I thought, but I expected the equation of the circle to contain only real terms :confused:
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    (Original post by The Silk Cobra)
    Yeah, that's what I thought, but I expected the equation of the circle to contain only real terms :confused:
    What do you mean? You now have a circle centred at (0,-1) with radius 2. This is completely real!
 
 
 
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