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    Please could someone explain part a) and part c) to me?

    For a) I understand that pi/2 implies a semi-circle, but could someone show from scratch how they would approach this question?

    Also, please could someone explain what's going in the graph for part c), because it looks nothing like a half line to me!

    Thanks
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    if you let z = x + iy

    (x + iy - 2i)/(x + iy + 2) can be multiplied by (x + 2 - iy)/(x + 2 - iy) and simplified to ax + iby.

    the arg of (p + iq) = pi/2, so tan pi/2 = q/p = tan (pi/2)

    so since tan ( pi/2 ) is undefined it follows that p = 0... from there you get the equation of a circle.
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    (Original post by the bear)
    if you let z = x + iy

    (x + iy - 2i)/(x + iy + 2) can be multiplied by (x + 2 - iy)/(x + 2 - iy) and simplified to ax + iby.

    the arg of (p + iq) = pi/2, so tan pi/2 = q/p = tan (pi/2)

    so since tan ( pi/2 ) is undefined it follows that p = 0... from there you get the equation of a circle.
    Thanks for the reply!

    I'm sorry but I still don't understand. I have no idea how to construct an answer to the question. I.e. I'm not sure why the semi-circle is in the quadrant it's in.
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    Zacken
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    (Original post by PhyM23)
    Zacken
    You know that the \arg(x+iy) is given by \arctan \frac{y}{x} (with some restrictions that we'll talk about later) we then have:

    \displaystyle \arg \frac{x+i(y-2)}{(x+2) + iy} = \arctan \frac{x+ i(y-2)}{(x+2) + iy}, rationalise this - the denominator will be ((x+2) + iy)(x+2 - iy).

    But since you know: \arctan \lambda = \frac{\pi}{2} \Rightarrow \lambda = \tan \frac{\pi}{2} = \infty (unrigorously) where \lambda  = the rationalised thing.

    But the only way you can have something be infinity is if the denominator is zero. So you know that ((x+2) + iy)(x+2 - iy) = 0 which is the equation of a circle.
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    (Original post by Zacken)
    You know that the \arg(x+iy) is given by \arctan \frac{y}{x} (with some restrictions that we'll talk about later) we then have:

    \displaystyle \arg \frac{x+i(y-2)}{(x+2) + iy} = \arctan \frac{x+ i(y-2)}{(x+2) + iy}, rationalise this - the denominator will be ((x+2) + iy)(x+2 - iy).

    But since you know: \arctan \lambda = \frac{\pi}{2} \Rightarrow \lambda = \tan \frac{\pi}{2} = \infty (unrigorously) where \lambda  = the rationalised thing.

    But the only way you can have something be infinity is if the denominator is zero. So you know that ((x+2) + iy)(x+2 - iy) = 0 which is the equation of a circle.
    Thanks for replying

    That makes sense, but when I expand ((x+2) + iy)(x+2 - iy) = 0 I get (x+2)^2 + y^2 = 0, but the radius can't be zero?

    And how does finding the equation of the circle help with knowing which quadrant to put the semi-circle in, and where you have to draw the half-lines to show where the angles make pi/2?
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    (Original post by PhyM23)
    ...
    Lol. ignore what I said. :lol: We have \frac{x + i(y-2)}{x+2 + iy}, rationalise this and then separate it into real and imaginary parts. Let's call the imaginary part m and the real part n.

    Then you know that \arctan \frac{m}{n} = \frac{\pi}{2} \Rightarrow n = 0. This will get you the equation of a circle.

    Now you know that you need m , n > 0 - so that gets you the restrictions that forces it to be only a semi-circle.
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    (Original post by Zacken)
    Lol. ignore what I said. :lol: We have \frac{x + i(y-2)}{x+2 + iy}, rationalise this and then separate it into real and imaginary parts. Let's call the imaginary part m and the real part n.

    Then you know that \arctan \frac{m}{n} = \frac{\pi}{2} \Rightarrow n = 0. This will get you the equation of a circle.

    Now you know that you need m , n > 0 - so that gets you the restrictions that forces it to be only a semi-circle.


    Ah I think I understand part a) now, except I'm not sure how m , n > 0 gets the restrictions for a semi-circle...

    Also, why have they in the solutions I posted not just done a single half line for part c)? The graph looks like a mess!
 
 
 
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