What is the Cauchy Integral Theorem and How Does it Apply to Complex Numbers?

In summary, using Cauchy's Integral Theorem, it is shown that the whole complex has a zero following result for the three given integrals, as all of them have singular points inside their contours. This is further supported by the Residue Theorem, which states that the integral is equal to 0 if the contour is closed and there are no singular points inside the contour. The integrals are then solved using the method of partial fractions, further proving the zero result.
  • #1
Ana
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Show that the whole complex has zero following result (Cauchy Integral Theorem):
 

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  • #2
Ana said:
Show that the whole complex has zero following result (Cauchy Integral Theorem):

NONE of these are 0 by Cauchy's Integral Theorem, which states that if your contour is closed and doesn't have any singular points in or on the contour, then the integral is equal to 0.

ALL of these have singular points inside their contours. You should use the Residue Theorem.
 
  • #3
Ana said:
Show that the whole complex has zero following result (Cauchy Integral Theorem):

Wellcome on MHB Ana!...

According to the Residue Theorem is $\displaystyle \int_{C} f(z)\ d z = 2\ \pi\ i\ \sum_{k} r_{k}$, where the $r_{k}$ are the residues of f(*) inside C...

First integral: $\displaystyle \int_{|z|=2} \frac{d z}{z^{2}-1}\ (1)$

Here f(*) has two poles in z=1 and z=-1, both inside C. Is...

$\displaystyle r_{1} = \lim_{z \rightarrow 1} (z-1)\ f(z) = \frac{1}{2}$

$\displaystyle r_{-1} = \lim_{z \rightarrow - 1} (z+1)\ f(z) = - \frac{1}{2}$

... so that...

$\displaystyle \int_{|z|=2} \frac{d z}{z^{2}-1} = 2\ \pi\ i\ (r_{1} + r_{-1})=0\ (2)$

Second integral: $\displaystyle \int_{|z|=2} \frac{d z}{z^{2}+1}\ (3)$

Here f(*) has two poles in z=i and z=-i, both inside C. Is...

$\displaystyle r_{i} = \lim_{z \rightarrow i} (z-i)\ f(z) = \frac{1}{2\ i}$

$\displaystyle r_{-i} = \lim_{z \rightarrow - i} (z+i)\ f(z) = - \frac{1}{2\ i}$

... so that...

$\displaystyle \int_{|z|=2} \frac{d z}{z^{2}+1}= 2\ \pi\ i\ (r_{i} + r_{-i})=0 (4)$

Third integral: $\displaystyle \int_{|z|=2} \frac{d z}{z^{2} - (1-i)\ z -i}\ (5)$

Here f(*) has two poles in z=1 and z = -1 -i, both inside C. Is...

$\displaystyle r_{1} = \lim_{z \rightarrow 1} (z-1)\ f(z) = -\frac{1}{i}$

$\displaystyle r_{-1 - i} = \lim_{z \rightarrow 1 + i} (z + 1 + i)\ f(z) = \frac{1}{i}$

... so that...

$\displaystyle \int_{|z|=2} \frac{d z}{z^{2} - (1-i)\ z -i} = 2\ \pi\ i\ (r_{1} + r_{-1-i}) = 0\ (6) $

Kind regards

$\chi$ $\sigma$
 
  • #4
I thought the idea was to give the OP hints and assistance, not do his/her homework for them...
 
  • #5
chisigma said:
Wellcome on MHB Ana!...

According to the Residue Theorem is $\displaystyle \int_{C} f(z)\ d z = 2\ \pi\ i\ \sum_{k} r_{k}$, where the $r_{k}$ are the residues of f(*) inside C...

First integral: $\displaystyle \int_{|z|=2} \frac{d z}{z^{2}-1}\ (1)$

Here f(*) has two poles in z=1 and z=-1, both inside C. Is...

$\displaystyle r_{1} = \lim_{z \rightarrow 1} (z-1)\ f(z) = \frac{1}{2}$

$\displaystyle r_{-1} = \lim_{z \rightarrow - 1} (z+1)\ f(z) = - \frac{1}{2}$

... so that...

$\displaystyle \int_{|z|=2} \frac{d z}{z^{2}-1} = 2\ \pi\ i\ (r_{1} + r_{-1})=0\ (2)$

Second integral: $\displaystyle \int_{|z|=2} \frac{d z}{z^{2}+1}\ (3)$

Here f(*) has two poles in z=i and z=-i, both inside C. Is...

$\displaystyle r_{i} = \lim_{z \rightarrow i} (z-i)\ f(z) = \frac{1}{2\ i}$

$\displaystyle r_{-i} = \lim_{z \rightarrow - i} (z+i)\ f(z) = - \frac{1}{2\ i}$

... so that...

$\displaystyle \int_{|z|=2} \frac{d z}{z^{2}+1}= 2\ \pi\ i\ (r_{i} + r_{-i})=0 (4)$

Third integral: $\displaystyle \int_{|z|=2} \frac{d z}{z^{2} - (1-i)\ z -i}\ (5)$

Here f(*) has two poles in z=1 and z = -1 -i, both inside C. Is...

$\displaystyle r_{1} = \lim_{z \rightarrow 1} (z-1)\ f(z) = -\frac{1}{i}$

$\displaystyle r_{-1 - i} = \lim_{z \rightarrow 1 + i} (z + 1 + i)\ f(z) = \frac{1}{i}$

... so that...

$\displaystyle \int_{|z|=2} \frac{d z}{z^{2} - (1-i)\ z -i} = 2\ \pi\ i\ (r_{1} + r_{-1-i}) = 0\ (6) $

Kind regards

$\chi$ $\sigma$
Thank you very much for your help, but did not learn about the Theorem of Residues yet, so I think that needs to be solved using Cauchy.
 
  • #6
You can resolve issues using theorems of Cauchy and method
of partial fractions, so decided. Thank you all!
 
  • #7
Without disturbing Monsieur Cauchy You can proceed for the first integral as follows: because is...

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

... is also...

$\displaystyle \int_{C} f(z)\ dz = \frac{1}{2}\ \int_{C}\frac{d z}{z-1} - \frac{1}{2} \int_{C} \frac{d z} {z+1}= \frac{1}{2}\ \int_{C} \frac{d z}{1-z} - \frac{1}{2}\ \int_{C} \frac{d z}{1-z} = 0\ (2)$

The same of course for the second and third integral...

Kind regards

$\chi$ $\sigma$
 

Related to What is the Cauchy Integral Theorem and How Does it Apply to Complex Numbers?

What is the Cauchy Integral Theorem?

The Cauchy Integral Theorem is a fundamental theorem in complex analysis, named after the mathematician Augustin-Louis Cauchy. It states that if a function is complex differentiable within a closed contour, then the integral of that function along the contour is equal to zero.

What is the significance of the Cauchy Integral Theorem?

The Cauchy Integral Theorem is significant because it provides a powerful tool for evaluating complex integrals. It also establishes a connection between complex differentiability and the behavior of a function along a closed contour.

What are the conditions for the Cauchy Integral Theorem to hold?

The Cauchy Integral Theorem holds when the function is complex differentiable within the contour and the contour is a simple, closed curve. Additionally, the function must be continuous on and inside the contour.

How is the Cauchy Integral Theorem related to Cauchy's Formula?

Cauchy's Formula is a direct consequence of the Cauchy Integral Theorem. It states that if a function is complex differentiable within a contour, then the value of the function at any point within the contour can be calculated by integrating the function along the contour divided by 2πi.

What are the applications of the Cauchy Integral Theorem?

The Cauchy Integral Theorem has various applications in mathematics and physics. It is used to evaluate complex integrals, solve differential equations, and analyze the behavior of functions in complex analysis. It also has applications in fluid dynamics, electromagnetism, and quantum mechanics.

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