Question

a. Evaluate the integral ∫Γ1z dz from A=(−1,0) to B=(1,0) along the curve Γ, where (a) Γ is the upper half-circle |z|=1, Imz≥0; (b) Γ is the lower half-circle |z|=1, Imz≤0. b. Compute the integral of the function f(z)=z¯2 along the given curves Γ. part a) The piecewise linear path AOBC, where A=(−1,0), O=(0,0), B=(0,1), C=(1,0). part b) The segment AC.

Answer 1

a) For the upper half-circle Γ with |z|=1 and Imz≥0, the integral ∫Γ1z dz from A=(-1,0) to B=(1,0) evaluates to πi. For the lower half-circle Γ with |z|=1 and Imz≤0, the integral equals -πi.

b) For the piecewise linear path AOBC (A=(-1,0), O=(0,0), B=(0,1), C=(1,0)), the integral of the function f(z)=z¯2 is iπ/3. For the segment AC, the integral of f(z) results in -2i/3.

Step-by-step explanation:

a) When evaluating the integral along a curve in the complex plane, the upper half-circle Γ can be parameterized as z =
`e^{(it)` where t ranges from 0 to π. Substituting this into the integral yields ∫Γ1z dz = ∫₀
`^π e^{(it)` * i
`e^{(it)` dt = i * ∫₀
`^π e^{(2it)` dt. This integral over the upper half-circle evaluates to πi. Conversely, for the lower half-circle, the parameterization becomes z =
`e^{(it)` for t ranging from -π to 0 due to the change in the direction of traversal. This leads to the integral ∫Γ1z dz = -πi.

b) For the piecewise linear path AOBC, integrating f(z)=z¯2 requires evaluating the integral separately along AO, OB, and BC. The calculations yield iπ/3 for AO and -iπ/3 for OB, resulting in a net contribution of 2iπ/3. Along BC, the integral evaluates to -2i/3. Finally, for the segment AC, by combining the contributions from AO and BC, the integral of f(z) along AC results in -2i/3.

These calculations demonstrate the significance of path and orientation when computing complex line integrals and showcase how the chosen path impacts the resulting integral value. The use of different parameterizations and the understanding of contours in the complex plane are crucial in solving such integrals along varied paths.