There is no region r such that f is analytic on r and c is

Question

Question 1

F=y³i + (3xy² - 4)j

F=Pi+Qj

∂P/∂y=∂Q/∂x inD

∂P/∂y=∂/∂y=y³

∂P/∂y=3y²

∂Q/∂x=∂/∂x=(3xy²-4)

∂Q/∂x=3y²

∂P/∂y=∂Q/∂x inD

F is conservative.

2)
3)

Question 2:

1)
f(z)=2z/((z-i)) ;γ is the circle |z-i|=3

f(z)=2z/((z-i))

z=0 ; z=i

f is not analytic at z=i

z=iis inside C

Since |i-i|=0<3

There is no region R such that f is analytic on R and C is completely contained in R, so in this case Cauchy's theorem is not applicable

2)
f(z)=2z/((z-i))

∫_C 2z/((z-i)) = 2πi×[residue of 2z/((z-i)) at z=i ]

= 2πi×2(i)
= 4π

Question 3:

1)

First use the map f(z) = z². I figured that this would map the semi-disk onto the unit disk.

Transformation

f(z) = (z+1)/(z-1) t

O map this unit disk onto the left half-plane. Rotating this by 90° clockwise.

By this line of thought the mapping would be:

f(z) = (-i)(z²+1)/(z²-1)

f(z) = (z²+1)/(iz²-i)