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

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.