I have no idea what to do here..
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...
Not sure where to go from there.
Turn on thread page Beta
Argand diagram... watch
- Thread Starter
Last edited by galadriel100; 15-04-2006 at 06:24.
- 15-04-2006 06:21
- 15-04-2006 11:01
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!
- Thread Starter
- 15-04-2006 11:47
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...
- 15-04-2006 12:40
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.
- 15-04-2006 13:14
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.