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I want to solve the following :

(i) Show that $z^4+2z^2-z+1$ has exactly one root per plane quadrant.

My idea to prove (i) is by using Rouche theorem, by considering 4 cuts of the complex plane which are given as:

$$ \Gamma_{1}= \{z / Re(z)>0 \:\mbox{and}\: Im(z)>0 \}\\ \Gamma_{2}= \{z / Re(z)<0 \:\mbox{and}\: Im(z)>0 \}\\ \Gamma_{3}= \{z / Re(z)<0 \:\mbox{and}\: Im(z)<0 \}\\ \Gamma_{1}= \{z / Re(z)>0 \:\mbox{and}\: Im(z)<0 \}$$

So by having $f(z)=z^4+2z^2-z+1$ how can I choose g(z) in order to have proceed by Rouche theorem in each $\Gamma_{i}$ for $i=1,2,3,4$ Im justasking by an example in some quadrant in order to solve for the other quadrants . Also I was wondering if by solving for $\Gamma_{1}$ and $\Gamma_{2}$ we have the same for $\Gamma_{4}$ and $\Gamma_{3}$ respectively by some conjugate complex argument?Thanks:)

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  • $\begingroup$ Rouche's theorem require a simple closed contour. $\endgroup$ – Mohammad W. Alomari May 17 '16 at 22:42
  • $\begingroup$ So Rouche´s theorem seems not quite a good way to solve the problem? :P $\endgroup$ – fgp May 17 '16 at 22:45
  • $\begingroup$ Your choice is wrong, $\Gamma_i$, $i=1,2,3,4$ $\endgroup$ – Mohammad W. Alomari May 17 '16 at 22:51
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    $\begingroup$ Can you be more helpful @mwomath? $\endgroup$ – fgp May 17 '16 at 22:55
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In the view that $$ f(z)=(z^2+1)^2-z $$ is nearly a biquadratic expression, it makes sense to seek a nearby factorized expression of the form \begin{align} g(z)&=(z^2+c)^2-(az+b)^2=(z^2+c+az+b)(z^2+c-az-b) \\&=z^4+2cz^2+c^2-a^2z^2-2abz-b^2 \end{align} which then requires to satisfy $2c-a^2\approx 2$, $2ab\approx 1$, $c^2-b^2\approx 1$ as close as possible.

One may try $a=\frac23$ and thus $b=\frac34$, $c=1+\frac12a^2=\frac{11}9$, giving a constant term of $c^2-b^2=1+\frac{40}{81}-\frac{9}{16}=1+\frac{640-729}{16·81}=1-\frac{89}{1296}$ thus $$ g(z)=f(z)-\frac{89}{1296} $$ On the real axis, $$ f(x)=x^4+\frac32x^2+\frac12(x-1)^2+\frac12\ge\frac12>|f(x)-g(x)| $$ On the imaginary axis $$ |f(iy)|=\sqrt{(y^2-1)^4+y^2}\ge \max(|y|,|y^2-1|^2)>\frac1{4}>|f(iy)-g(iy)| $$ for $|y|>\frac{1}{4}$ or $|y|^2<\frac1{2}$, that is, for the whole imaginary axis. Thus both polynomial have the same number of roots in each of the quadrants. (If necessary, add segments of the circle of radius $2$ to the boundary, as $|f(z)|\ge(4-1)^2-2=7>|f(z)-g(z)|$ there.

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