for the circle mod.(z-1)=1 in argand plane prove arg(z-1)=2arg(z)?

Answer 1

Let arg z = theta (which I'll write as t). Write z = r cos t + r i sin t. Then ||(z-1)||^2 = 1 = (r cos t - 1)^2 + (r sin t)^2
=> r^2 cos^2 t - 2 r cos t + 1 + r^2 sin^2 t = 1
=> r^2 (1) - 2r cos t = 0
=> r = 2 cos t (r /= 0; the statement is true for r=0 by inspection)
=> z = 2 cos^2 t + 2i sin t cos t
= (cos 2t + 1) + i sin 2t
=> z - 1 = cos 2t + i sin 2t
=> arg(z-1) = 2t.
So arg(z-1) = 2 arg(z).