Evaluate C(2z+3)/z dz where C is upper half of the circle |z| = 2 in clockwise direction.

To solve this integral, we will use the method of residues. The function has a simple pole at z = 0.

Step 1: Find the Residue The residue of a function at a simple pole z = a can be found using the formula Res(f, a) = lim (z-a) * f(z) as z approaches a.

In this case, our function f(z) = (2z+3)/z and our pole a = 0. So, we have:

Res(f, 0) = lim (z-0) * (2z+3)/z as z approaches 0 = lim (2z^2+3z)/z as z approaches 0 = lim (2z + 3) as z approaches 0 = 2*0 + 3 = 3

Step 2: Use the Residue Theorem The Residue Theorem states that the integral of a function around a simple closed contour is 2πi times the sum of the residues of the function at its poles inside the contour.

In this case, we have only one pole inside our contour (the upper half of the circle |z| = 2), and we found its residue to be 3. So, we have:

∫C (2z+3)/z dz = 2πi * 3 = 6πi

However, since the contour is traversed in the clockwise direction, we need to take the negative of this result. So, the final answer is:

∫C (2z+3)/z dz = -6πi