Find the value of the integral (3z^3 + 2)/((z - 1)(z^2 + 9)) dz, taken counterclockwise around t

Name: Find the value of the integral (3z^3 + 2)/((z - 1)(z^2 + 9)) dz, taken counterclockwise around the circle (a) |z - 2| = 2; (b) |z| = 4.
Uploaded: 2023-07-01T01:29:31-08:00
Duration: 1 min 58 s
Channel: Madhur L
Description: Find the value of the integral (3z^3 + 2)/((z - 1)(z^2 + 9)) dz, taken counterclockwise around the circle (a) |z - 2| = 2; (b) |z| = 4.

Step 1: First, we need to factor the denominator of the integrand. We have (z - 1)(z^2 + 9) = z^

Question

Find the value of the integral \( \int_c \frac{3z^3 + 2}{(z - 1)(z^2 + 9)} dz \), taken counterclockwise around the circle (a) \(|z - 2| = 2\); (b) \(|z| = 4\).

          Find the value of the integral \( \int_c \frac{3z^3 + 2}{(z - 1)(z^2 + 9)} dz \), taken counterclockwise around the circle (a) \(|z - 2| = 2\); (b) \(|z| = 4\).