I managed to do this in terms of polar coordinates. How would one solve this using the fact that T(z) = z^2? (2) Consider the map T: D D, where D is the unit disk in the plane, given by T(r cos theta, r sin theta) = (r^2 cos 2theta,r2 sin 2theta), Using complex notation, z = x + iy, the map T can be written as T(z) = z^2, Show that the Jacobian determinant of T vanish only at the origin. Thus, away from the origin, T is locally one-to-one. However, show that T is not globally one-to- one on 2 minus the origin. Solution T(z) = T(rcost, rsint) Consider z^2 = (rcost, rsint).(rcost,rsint) = r^2cos^2t+r^2sin^2t = r^2 = (x^2+y^2) where x =rcost and y =rsint Thus T(z) = z^2 T(z) = z^2 =0 only when r=0 or at the origin. As T(z) = z^2, T(-z) = (-z)^2 =z^2 Thus z and -z have the same image under T. T is not one to one..

1. I managed to do this in terms of polar coordinates. How would one solve this using the fact that
T(z) = z^2? (2) Consider the map T: D D, where D is the unit disk in the plane, given by T(r cos
theta, r sin theta) = (r^2 cos 2theta,r2 sin 2theta), Using complex notation, z = x + iy, the map T
can be written as T(z) = z^2, Show that the Jacobian determinant of T vanish only at the origin.
Thus, away from the origin, T is locally one-to-one. However, show that T is not globally one-to-
one on 2 minus the origin.
Solution
T(z) = T(rcost, rsint)
Consider z^2 = (rcost, rsint).(rcost,rsint) = r^2cos^2t+r^2sin^2t
= r^2
= (x^2+y^2) where x =rcost and y =rsint
Thus T(z) = z^2
T(z) = z^2 =0 only when r=0 or at the origin.
As T(z) = z^2, T(-z) = (-z)^2 =z^2
Thus z and -z have the same image under T.
T is not one to one.