Let $f(z)=\frac{1}{z^2+6z+9}$ defined in the complex plane. The integral ∮${}_c$ f(z)dz over the contour of a circle c with center at the origin and unit radius is ___.

In the range: 0 - 0

Concept:

Cauchy integral formula:

If f(z) is analytic within a closed curve and if a is any point within C (contour), then

f(a) = $\frac{1}{2\pi i}\oint \frac{f(z)dz}{z-a}$

Calculation:

Given

$f(z)=\frac{1}{z^2+6z+9}$

Contour is a unit radius circle with center is the origin

$f(z)=\frac{1}{(z+3{)}^2}$

Pole of f(z) is -3, -3 and both poles are outside of the given unit circle (it is shown in below fig.)



Here all-poles of f(z) outside the circle so

**$\oint f(z)dz=0$ **