Question

Let \( f(z) = \dfrac{1}{z^2 + 6z + 9} \) defined in the complex plane. The integral \( \oint_{c} f(z) \, dz \) over the contour of a circle \( c \) with center at the origin and unit radius is \(\underline{\hspace{2cm}}\).
 

Hint:

If the contour does not enclose any singularities, the integral of the function around that contour is zero.

Answer 1

Correct Answer: 0

The function \( f(z) = \frac{1}{z^2 + 6z + 9} \) has a pole at \( z = -3 \), which is outside the contour of the unit circle centered at the origin. Since the contour does not enclose the singularity, by Cauchy's integral theorem, the integral is zero. Thus: \[ \oint_{c} f(z) \, dz = 0 \] Thus, the value of the integral is \( 0 \).