Question

Evaluate \[ \oint_C \frac{z+2}{z^2+2z+2}\,dz, \] where \(C\) is the circle \[ |z-\alpha|=1, \qquad \alpha=-1+\frac{3}{2}i. \]

• A. \(0\)
• B. \(2\pi i\)
• C. \(\pi(1+i)\)
• D. \(\pi(2-i)\)

Hint:

For contour integrals of rational functions, first locate all poles, determine which poles lie inside the contour, then apply the Residue Theorem.

Answer 1

The Correct Option is B

Concept: To evaluate contour integrals of analytic functions, we use Cauchy's Integral Theorem and the Residue Theorem. The integral depends on the poles enclosed by the contour.

Step 1: Factorize the denominator.
\[ z^2+2z+2 = (z+1)^2+1. \] Therefore, \[ z^2+2z+2 = (z+1-i)(z+1+i). \] Hence the poles are \[ z=-1+i, \qquad z=-1-i. \]

Step 2: Check which pole lies inside the contour.
The contour is \[ |z-(-1+\tfrac32 i)|=1. \] Distance of \(z=-1+i\) from the center: \[ \left|(-1+i)-\left(-1+\frac32 i\right)\right| = \frac12<1. \] Hence \(z=-1+i\) lies inside. Distance of \(z=-1-i\): \[ \left|(-1-i)-\left(-1+\frac32 i\right)\right| = \frac52>1. \] Hence \(z=-1-i\) lies outside.

Step 3: Compute the residue at the enclosed pole.
\[ \text{Res}_{z=-1+i} \frac{z+2}{(z+1-i)(z+1+i)} = \left. \frac{z+2}{z+1+i} \right|_{z=-1+i}. \] \[ = \frac{1+i}{2i}. \] Multiplying numerator and denominator by \(-i\), \[ = \frac{(1+i)(-i)}{2} = \frac{1-i}{2}. \]

Step 4: Apply the Residue Theorem.
\[ \oint_C \frac{z+2}{z^2+2z+2}\,dz = 2\pi i \left(\frac{1-i}{2}\right). \] \[ = \pi i(1-i). \] \[ = \pi(1+i). \] Hence, \[ \boxed{\pi(1+i)} \]