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    • Thread Starter

    I have no idea what to do here..:confused:

    On an Argand diagram draw vectors representing z and w
    where 0<argz<pi/2 and -pi/2<argw<0

    Hence, show clearly on your Argand diagram the following stating whether you are reffering to a vector, a length or an angle.
    a) -z (b) conjugate of z (c)z-w ...etc etc...

    So, with the first part, would I just shade in the area between 0 and pi/2 and label it as arg z?

    Another thing I don't understand is;
    Let z=x+iy, where x and y are real numbers.
    a) show that there are 3 non-zero complex numbers that satisfy the equation conjugate or z=z^2
    I expanded it out, and got...
    x-iy=x^2 +2iyx+y^2
    Not sure where to go from there.


    I've attached an image of how your argand diagram should look. Things to note are:

    1) arg(z) is simply the angle the vector to the complex number subtends ANTI-CLOCKWISE from the +ve x-axis. So if arg(z) is between 0 and pi/2. All this means is that the complex number has to lie in the top-right quadrant. So you can put it anywhere in +ve x, +ve y region.

    2) arg(w) is between -pi/2 and 0, so it has to be in the lower-right quadrant of +ve x, -ve y. It's a negative argument because this angle is being measured clockwise from the x-axis.

    3) Conjugate of z is z*. So if z = x + iy, z* = x - iy i.e. reflect it in the x-axis.

    4) Negative of z is -z. So just make x+iy into -x-iy.

    5) When adding complex numbers, use the head-to-tail rule for adding vectors z+w, and just add their components. When subtracting, use the head-to-tail rule again, but this time do z+(-w).

    6) The magnitude of a vector |z| is simply it's length, sqrt(x^2 + y^2).

    I don't understand the second part of your question though. Are you sure you've typed it correctly?

    z* = z^2 means the conjugate of z equals a number with double z's length and twice z's argument. But by definition |z| = |z*| and arg(z*) = -arg(z), so I don't really see any solution, let alone three!
    Attached Images
    • Thread Starter

    Thanks so much for your help Worzo! You even went to the trouble of drawing a diagram!

    I'm pretty sure I typed the secodn question correctly. But here's what the question actually looks like anyway, just incase...
    Attached Images

    For z = x +iy the equation z* = z^2 is equivalent to

    x - iy = x^2 + 2ixy - y^2

    Equate real and imaginaries

    x = x^2 - y^2
    2xy = -y

    Solve to give four solutions

    x = 0, y = 0
    x = 1, y = 0
    x = -1/2, y = sqrt(3)/2
    x = -1/2, y = -sqrt(3)/2

    three of which give rise to nonzero z.

    Of course, mikesgt2 - I thought z^2 meant zz*, and so got confused because the question was asking z*, a complex number = zz*, a real length. Having a mare.
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Updated: April 15, 2006
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