Question

2. Find the image of the upper half-plane by the mapping 1-χα 1 za where 0< α< 1 and 20 has its principal value.

Answer 1

We made equation (2) by using De-Movries theorem and then took its imaginary part...thank you so much!

W = 1 - z^alpha/1 + z^a implies W + z^alpha w = 1 - z^alpha implies z^alpha (1 + w) = 1 - w implies z^alpha = (1 - w)/(1 + w) implies z = (1 - w/1 + w)^1/alpha Now upper half plane means in (z) > 0 implies in (z) = (1 - w/1 + w)^1/alpha > 0 implies (1 - u - I v/1 + u + iv)^1/alpha > 0 where w = u + I v implies (1 - u - I v/1 + u + I v times 1 + u - iv/1 + u - I v)^1/alpha > 0 implies In (1 - u^2 - v^2 - I (uv - uv + v)/(1 + u^2 + v^2)^1/alpha > 0 implies In (1 - u^2 - v^2 - I v)^1/alpha > 0 let 1 - u^2 - v^2 = r cos theta u - v = r sin theta then In (1 - u^2 - v^2 - I v) = -v = r sin theta equation 1 & In (1 - u^2 - v^2 - I v)^1/alpha = In (r) \r\n