How do I evaluate the contour integral for $f'(z)/f(z)$ around $|z|=4$?

[Fermat1]

Let $f(z)=\frac{(z^2+1)^2}{(z^2+2z+2)^3}$ . Evaluate the contour integral of $f'(z)/f(z)$ around the circle $|z|=4$?

How do I do this without having to find $f'(z)$?

Thanks

 

[chisigma]

Let $f(z)=\frac{(z^2+1)^2}{(z^2+2z+2)^3}$ . Evaluate the contour integral of $f'(z)/f(z)$ around the circle $|z|=4$?

How do I do this without having to find $f'(z)$?

Thanks

An immediate answer is given by what I consider one of the most beautiful theorems of complex analysis: the principle of the argument ...

Argument Principle -- from Wolfram MathWorld

Fermat thank you for giving me a good procedure to be followed in a mathematical note that I'm carrying on about the logarithm function in the complex domain ...

Kind regards

$\chi$ $\sigma$

 

[Fermat1]

An immediate answer is given by what I consider one of the most beautiful theorems of complex analysis: the principle of the argument ...

Argument Principle -- from Wolfram MathWorld

Fermat thank you for giving me a good procedure to be followed in a mathematical note that I'm carrying on about the logarithm function in the complex domain ...

Kind regards

$\chi$ $\sigma$

Ok, the number of roots is 2. How about the poles? Thanks

 

[Fermat1]

An immediate answer is given by what I consider one of the most beautiful theorems of complex analysis: the principle of the argument ...

Argument Principle -- from Wolfram MathWorld

Fermat thank you for giving me a good procedure to be followed in a mathematical note that I'm carrying on about the logarithm function in the complex domain ...

Kind regards

$\chi$ $\sigma$

$f$ has 2 poles (at -1+i and -1-i) so the integral is 2-2=0. Correct? Can you tell me how poles a polynomial has.

 

[Opalg]

$f$ has 2 poles (at -1+i and -1-i) so the integral is 2-2=0. Correct? Can you tell me how poles a polynomial has.

Read chisigma's link to Wolfram MathWorld more carefully! The zeros and poles must be counted according to their multiplicities.

 

[Fermat1]

Read chisigma's link to Wolfram MathWorld more carefully! The zeros and poles must be counted according to their multiplicities.

so there are 4 roots and 2 poles of order 3?

 

[chisigma]

You have to apply the relation...

$\displaystyle \int_{\gamma} \frac{f^{\ '}(z)}{f(z)}\ dz = 2\ \pi\ i\ (n-m)\ (1)$

... where...

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

... and $\gamma$ is the circle for which $|z|=4$. If all the roots of p(*) and q(*) fall inside $\gamma$ [that is the case...], then n is the degree of p(*) and m is the degree of q(*), so that (Speechless) ...

Kind regards

$\chi$ $\sigma$