Question

The value of the integral
\[ \int_C \frac{6z - 5}{z^2 + 4z + 5} \, dz, \text{where } C \text{ is the circle } |z| = 1, \text{ is ............} \]

• A. 0
• B. $2\pi i$
• C. $\pi$
• D. $\frac{i\pi}{2}$

Hint:

When using the Residue Theorem, ensure that the poles of the function are inside the contour. If there are no poles inside the contour, the integral is zero.

Answer 1

The Correct Option is A

We are given the integral: \[ \int_C \frac{6z - 5}{z^2 + 4z + 5} \, dz \] where $C$ is the circle $|z| = 1$. The integral is a contour integral, so we can apply the Residue Theorem to evaluate it.
1. Step 1: Factor the denominator
The denominator is: \[ z^2 + 4z + 5 = (z + 2)^2 + 1 \] This is a quadratic expression, and we can see that it has no real roots. The roots are complex and occur at: \[ z = -2 \pm i \] 2. Step 2: Check if the poles are inside the contour
The contour $C$ is the circle $|z| = 1$. We need to check if the poles $z = -2 + i$ and $z = -2 - i$ lie inside the contour.
The distance of each pole from the origin is: \[ |z - (-2 + i)| = \sqrt{(-2)^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} (\text{which is greater than 1}) \] \[ |z - (-2 - i)| = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} (\text{which is greater than 1}) \] Both poles are outside the contour, as their distances from the origin are greater than 1. 3. Step 3: Apply the Residue Theorem
Since both poles lie outside the contour, the integral does not enclose any singularities. According to the Residue Theorem, the integral of a function around a closed curve that does not enclose any singularities is zero. Therefore, the value of the integral is: \[ \boxed{0} \] Thus, the value of the integral is $0$.