Why is the integral of $\frac{1}{z+2}$ over $\gamma(0;1)$ equal to zero?

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Discussion Overview

The discussion centers on the evaluation of the integral $$\int_{\gamma(0;1)}\frac{1}{z+2} \mathrm{d}z$$ and why it equals zero. Participants explore concepts related to complex analysis, specifically Cauchy's Integral Theorem and the implications of singularities in relation to contour integrals.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant notes that the point $$z=-2$$ lies outside the contour $$\gamma(0;1$$, suggesting that this is key to understanding why the integral is zero.
  • Another participant agrees, stating that since the pole lies outside the closed curve, the integral evaluates to zero.
  • A different participant references Cauchy's Integral Theorem, explaining that the function is holomorphic everywhere on and inside the contour, thus supporting the conclusion that the integral is zero.
  • One participant elaborates on the fundamental result regarding integrals around poles, indicating that if the pole were included within the contour, the integral would yield a non-zero result.
  • Another participant reiterates the application of Cauchy's Theorem, affirming that the integral equals zero because the only pole is outside the circular curve.

Areas of Agreement / Disagreement

Participants generally agree that the integral is zero due to the pole being outside the contour. However, there are variations in the explanations provided, with some emphasizing Cauchy's Theorem while others focus on the nature of poles in relation to the contour.

Contextual Notes

Some participants reference fundamental results and theorems without fully detailing the conditions under which they apply, which may lead to assumptions about the audience's familiarity with complex analysis concepts.

shen07
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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?
 
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Re: Reason why Integral Zero

Hi shen07! :)

shen07 said:
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 $2\pi i$.
 
Re: Reason why Integral Zero

shen07 said:
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.

Proof of the Theorem
 
Re: Reason why Integral Zero

shen07 said:
Well i know by a fundamental result that:
$$\int_{\gamma(a;r)}\frac{1}{z-a} \mathrm{d}z=2\pi\imath$$

By using $$\gamma(a;r)$$ you are making sure that the pole at $$z=a$$ is included in your curve by letting it the center of the disk since the radius is nonzero .
 
Re: Reason why Integral Zero

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
$$\int_{\gamma(0;1)}\frac{1}{z+2} \mathrm{d}z=0$$
 
Re: Reason why Integral Zero

shen07 said:
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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