Let S₁ = z C : ( z - 2 z + 2 ) = 4 and S₂ = z C : ( z - 2 z + 2 ) = 3 4 . The area of the region enclosed by the curves S₁ and S₂ is

Step-by-step solution

Let w = z - 2 z + 2 . Substituting z = x + iy , we have: w = (x - 2) + iy (x + 2) + iy = x^2 + y^2 - 4 + i(4y) (x + 2)^2 + y^2 For S₁ , (w) = 4 , so Re (w) = Im (w) > 0 . x^2 + y^2 - 4 = 4y x^2 + (y - 2)^2 = 8 Since Im (w) > 0 , we have 4y > 0 y > 0 . Thus, S₁ is the major arc of the circle x^2 + (y - 2)^2 = 8 in the upper half-plane, connecting (-2, 0) and (2, 0) . For S₂ , (w) = 3 4 , so Re (w) = - Im (w) 0 . x^2 + y^2 - 4 = -4y x^2 + (y + 2)^2 = 8 Since Im (w) > 0 , we have y > 0 . Thus, S₂ is the minor arc of t