complex plane and functions!!!

Denote C be the set of all complex numbers .

1. Let z = x + yi

(a) z^2 = (x+yi)^2 = (x^2 - y^2) + 2xyi
Im(z^2) > 0 => 2xy > 0 => ( x > 0 and y > 0 ) or ( x < 0 and y < 0 )
So , the region = { z in C | (Rez and Imz > 0) or ( Rez and Imz < 0 ) }

(b) |z^2| < 1 => |z|^2 < 1 => (|z| - 1)(|z| + 1) < 0
=> -1 < |z| < 1 => 0 <= |z| < 1 ( since |z| >= 0 for any complex number z )
So , the region = { z in C | 0<=|z| < 1 }

2. Let z = x+yi , denote z'=x-yi be the complex conjugate of z and
write r=|z|=sqrt( x^2 + y^2)

(a) e^(1/z) = e^(z'/|z|^2)=e^[x/r^2 -i( y/r^2)]
Im(e^(1/z)) = 0 => e^(x/r^2)sin(-y/r^2) = 0
so , sin(y/r^2) = 0 . y/r^2 = npi , where n is an integer .
If n = 0 , then y = 0 ( but x =/= 0 , otherwise , 1/z is undefined )
In this case , the curve is the straight line y = 0 excluding (0,0)
If n =/= 0 , then y/npi = x^2 + y^2
x^2 + y^2 - (1/npi)y + (1/2npi)^2 = (1/2npi)^2
x^2 + (y - 1/2npi)^2 = (1/2npi)^2
It is a family of circles center at (0,1/2npi) with radius 1/2npi , where n is
non-zero integers .

(b) Similar to (a) , we have cos(x/r^2) = 0 => x/r^2 = 2npi+/-(pi/2)
Let K_n = 2npi+/-(pi/2)
we have x^2 + y^2 = (1/K_n)x
( x - 1/2K_n)^2 + y^2 = (1/2K_n)^2
It is a family of circles center at (1/2K_n , 0) with radius 1/2K_n
Where K_n = 2npi+/- (pi/2) and n is an integer

2012-03-22 09:17:30 補充：
(b) 更正 :
similar to (a) , we have cos(y/r^2) = 0 => y/r^2 = 2npi+/-(pi/2)
Let K_n = 2npi+/-(pi/2)
we have x^2 + y^2 = (1/K_n)y
x^2 + ( y - 1/2K_n)^2 = (1/2K_n)^2
It is a family of circles center at (0,1/2K_n) with radius 1/2K_n
Where K_n = 2npi+/- (pi/2) and n is an integer