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

[Ana]

Show that the whole complex has zero following result (Cauchy Integral Theorem):

 

[Prove It]

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.

 

[chisigma]

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

$\chi$ $\sigma$

 

[Nono713]

I thought the idea was to give the OP hints and assistance, not do his/her homework for them...

 

[Ana]

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.

 

[Ana]

You can resolve issues using theorems of Cauchy and method
of partial fractions, so decided. Thank you all!

 

[chisigma]

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

$\chi$ $\sigma$