# Thread: Reason why Integral is Zero

1. Hello i would like to know why this integral is Zero:

$$\int_{\gamma(0;1)}\frac{1}{z+2} \mathrm{d}z$$

Well i know by a fundamental result that:
$$\int_{\gamma(a;r)}\frac{1}{z-a} \mathrm{d}z=2\pi\imath$$

But here the point $$z=-2$$ lies outside $$\gamma(0;1)$$ so what is the reasoning behind could you help me please?

2.

3. Hi shen07!

Originally Posted by shen07
But here the point $$z=-2$$ lies outside $$\gamma(0;1)$$ so what is the reasoning behind could you help me please?
I believe you already said it: the "pole" lies outside the closed curve.
In that case the circular integral is zero.
If the curve were around a pole, it would be $2\pi i$.

4. Originally Posted by shen07
Hello i would like to know why this integral is Zero:

$$\int_{\gamma(0;1)}\frac{1}{z+2} \mathrm{d}z$$

Well i know by a fundamental result that:
$$\int_{\gamma(a;r)}\frac{1}{z-a} \mathrm{d}z=2\pi\imath$$

But here the point $$z=-2$$ lies outside $$\gamma(0;1)$$ so what is the reasoning behind could you help me please?
Cauchy's Integral Theorem states that for any complex function which is closed and holomorphic everywhere in and on the boundary, its contour integral is equal to 0. As your function only has a singular point at z = -2, which is not in the boundary, your function satisfies the conditions and so Cauchy's Integral Theorem applies.

5. Originally Posted by shen07

Well i know by a fundamental result that:
$$\int_{\gamma(a;r)}\frac{1}{z-a} \mathrm{d}z=2\pi\imath$$
By using $\displaystyle \gamma(a;r)$ you are making sure that the pole at $\displaystyle z=a$ is included in your curve by letting it the center of the disk since the radius is nonzero .

What if i use Cauchy's Theorem, since $$f(z)=\frac{1}{z+2}$$ is holomorphic on and inside $$\gamma(0;1)$$, using Cauchy's Theorem
$$\int_{\gamma(0;1)}\frac{1}{z+2} \mathrm{d}z=0$$

7. Originally Posted by shen07
What if i use Cauchy's Theorem, since $$f(z)=\frac{1}{z+2}$$ is holomorphic on and inside $$\gamma(0;1)$$, using Cauchy's Theorem
$$\int_{\gamma(0;1)}\frac{1}{z+2} \mathrm{d}z=0$$
Yes , since the only pole is at $z=-2$ which is out of the circular curve .

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