Evaluate the line integrals ∫𝐶 𝑓 (𝑧)𝑑𝑧 along the curve 𝐶, where(a) 𝑓 (𝑧) = 1𝑧 − 2 , 𝐶 : |𝑧| = 3 in the counterclockwise direction.(b) 𝑓 (𝑧) = 𝑧3 , 𝐶 : |𝑧 − 𝑖| = 2 in the counterclockwise direction.(c) 𝑓 (𝑧) = 1𝑧2 + 1 , 𝐶 : |𝑧| = 1 in the counterclockwise direction.(d) 𝑓 (𝑧) = |𝑧|2 , 𝐶 : 𝑡2 + 𝑖𝑡 where 0 ≤ 𝑡 ≤ 1.

Sure, let's solve these line integrals one by one.

(a) The function f(z) = 1/(z-2) has a singularity at z = 2. However, the curve C: |z| = 3 does not enclose this singularity. Therefore, by Cauchy's integral theorem, the line integral is zero.

(b) The function f(z) = z^3 is entire, meaning it is analytic everywhere in the complex plane. The curve C: |z - i| = 2 is a circle centered at z = i with radius 2. By Cauchy's integral theorem, the line integral of an entire function over any closed curve is zero.

(c) The function f(z) = 1/(z^2 + 1) has singularities at z = i and z = -i. The curve C: |z| = 1 is a circle centered at the origin with radius 1. This curve encloses the singularity at z = i but not the one at z = -i. Therefore, by the residue theorem, the line integral is 2πi times the residue at z = i. The residue at z = i is lim_(z->i) (z-i)f(z) = lim_(z->i) (z-i)/(z^2 + 1) = 1/2. Therefore, the line integral is 2πi * 1/2 = πi.

(d) The function f(z) = |z|^2 is not analytic, so we cannot use Cauchy's integral theorem or the residue theorem. Instead, we need to parameterize the curve and compute the line integral directly. The curve C: t^2 + it for 0 ≤ t ≤ 1 is a parabola in the complex plane. We can parameterize this curve as z(t) = t^2 + it, dz/dt = 2t + i, 0 ≤ t ≤ 1. Then the line integral is ∫_C f(z) dz = ∫_0^1 f(z(t)) dz/dt dt = ∫_0^1 |t^2 + it|^2 (2t + i) dt. This integral needs to be computed directly.