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- Thread starter dwsmith
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- Jan 29, 2012

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http://www.fernandorevilla.es/docen...o-del-argumento-ceros-en-regiones-no-acotadas

Ask if you have any doubt.

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- Feb 13, 2012

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According to the so called The 'argument principle', if f(*) is analytic in D and $\gamma$ is the 'frontier' of D, then the number of zeroes of f(*) in D is given by...For a fixed number a, find the number of solutions $z^5+2z^3-z^2+z=a$ satisfying $\text{Re} \ z>0$.

Not to sure on how to tackle this one.

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

We consider $\displaystyle f(z)=z^{5}+2\ z^{3} -z^{2}+z-a$ and we set p the number of roots of f(*) with positive real part and q the number of roots of f(*) with negative real part. Of course is p+q=5. Now if we apply (1) and choose D as the 'big half circle tending to the left half plane' we obtain...

$\displaystyle q= \frac{1}{2\ \pi} \int_{- \infty}^{+ \infty} \frac{f^{'}(i\ y)}{f(i\ y)}\ dy = \frac{1}{2\ \pi} \int_{- \infty}^{+ \infty} \frac {(5\ y^{4} -6\ y^{2} +1) -2\ i\ y}{(y^{2}-a) +i\ (y^{5}-y^{3}+y)}\ dy$ (2)

The detail of computation of integral (2) are however 'a little complex' and that is 'postposed' to a successive post...

Kind regards

$\chi$ $\sigma$

- Jan 29, 2012

- 661

Denote $p(z)=z^5+2z^3-z^2+z-a$. Then, $p(z)=z^5\left(1+\dfrac{2}{z^2}-\dfrac{1}{z^3}+\dfrac{1}{z^4}-\dfrac{a}{z^5}\right)$ if $z\neq 0$.$

z^5\left(1+\frac{2}{z^2}-\frac{1}{z^3}+\frac{1}{z^4}-\frac{a}{z^5}\right)

$ After you did this, what did you do for the argument? I don't understand.

Using a well known property of the argument,

$\arg p(z)=5\arg z+\arg \left(1+\dfrac{2}{z^2}-\dfrac{1}{z^3}+\dfrac{1}{z^4}-\dfrac{a}{z^5}\right) $

Consider the curve $\gamma (t)=Re^{it}\;\; (t\in [-\pi/2,\pi/2])$. The increment $\Delta \arg z$ of the argument of $z$ in this curve is $\pi$. For $R\to +\infty$ and due to the continuity of the argument, we have

$\Delta_{z\in \gamma}\arg p(z)\to 5\pi +0+0+0+0+0=5\pi$.

- Feb 13, 2012

- 1,704

The integral (2) supplies in fact the 'formal solution' of the proposed question... but of course its approach is 'a little unpleasant'... a more simple way to arrive to the goal is considering that the (2) represents the global 'rotation' for y going from minus to plus infinity of the function...According to the so called The 'argument principle', if f(*) is analytic in D and $\gamma$ is the 'frontier' of D, then the number of zeroes of f(*) in D is given by...

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

We consider $\displaystyle f(z)=z^{5}+2\ z^{3} -z^{2}+z-a$ and we set p the number of roots of f(*) with positive real part and q the number of roots of f(*) with negative real part. Of course is p+q=5. Now if we apply (1) and choose D as the 'big half circle tending to the left half plane' we obtain...

$\displaystyle q= \frac{1}{2\ \pi} \int_{- \infty}^{+ \infty} \frac{f^{'}(i\ y)}{f(i\ y)}\ dy = \frac{1}{2\ \pi} \int_{- \infty}^{+ \infty} \frac {(5\ y^{4} -6\ y^{2} +1) -2\ i\ y}{(y^{2}-a) +i\ (y^{5}-y^{3}+y)}\ dy$ (2)

The detail of computation of integral (2) are however 'a little complex' and that is 'postposed' to a successive post...

$\displaystyle \frac{1}{2\ \pi\ i}\ \text{arg}\ f(i\ y) = \frac{1}{2\ \pi\ i}\ i\ \tan^{-1} \frac{y^{5}-y^{3}+y}{y^{2}-a}$ (3)

Observing (3) we note that the numerator doesn't depends from a and its contribution is in any case $5\ \pi$... but the denominator depends from a and it doesn't change sign for $a<0$ and change sign two times for $a>0$, so that we conclude that is $q=\frac{5+1}{2}=3 \implies p=2$ for $a<0$ and $q=\frac{5-1}{2}=2 \implies p=3$ for $a>0,\ a \ne 1$. For $a=0$ is $p=q=2$ and for $a=1$ is $p=1,\ q=2$...

Kind regards

$\chi$ $\sigma$

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