Originally Posted by

**ymar** I have to find out how many roots the polynomial

$\displaystyle f(z)=z^8+3z^7-z^5+2z^4+z+1,\,z\in\mathbb{C}$

has in the first quadrant.

Certainly, I need to use the argument principle, but I don't see how. Here's what I tried.

All zeroes of $\displaystyle f$ lie in $\displaystyle \{z\in\mathbb{C}:|z|<4\},$ which follows from applying the Cauchy bound. Suppose I can prove that $\displaystyle f$ has no real and no imaginary roots. Then, since the function is holomorphic, I have to calculate

$\displaystyle \frac{1}{2\pi i}\left (\int_{I_1}\frac{f'}{f}dz+\int_{I_2}\frac{f'}{f}dz +\int_{D}\frac{f'}{f}dz\right ),$

where

$\displaystyle I_1=\{z:0\leq Re(z)\leq 4,\,Im(z)=0\}$

$\displaystyle I_2=\{z:0\leq Im(z)\leq 4,\,Re(z)=0\}$

$\displaystyle D=\{z:0\leq Im(z),\,0\leq Re(z),\,(Re(z))^2+(Im(z))^2=16\},$

(integrals have to be calculated in the appropriate directions of course).

But to calculate those integrals, don't I need to know the zeroes of $\displaystyle f?$ Only then would I be able to integrate by residues, right? It's supposed to be very simple.