Prove that \(f(z) = z^4 + 2z^3 -2z + 10\) has exactly one root in each open quadrant.
Solution
First exclude roots on the coordinate axes. For real \(x\), \[ f(x)=(x^2+x-1)^2+x^2+9>0. \] For real \(t\), \[ f(it)=t^4+10-2i(t^3+t), \] whose real part is strictly positive. Thus \(f\) has no zeros on either axis.
Now count zeros in the first quadrant. Let \(\Gamma_R\) be the positively oriented boundary of the quarter-disk \[ \{z:|z|<R,\ 0<\arg z<\pi/2\}. \] Along the positive real edge, \(f\) is positive, so the change of argument is \(0\).
On the circular edge \(z=Re^{it}\), \(0\le t\le\pi/2\), \[ \frac{f(z)}{z^4} =1+\frac2z-\frac2{z^3}+\frac{10}{z^4} \longrightarrow1 \] uniformly as \(R\to\infty\). Hence the change of argument of \(f\) along this arc tends to the change of argument of \(z^4\), namely \[ 4\cdot\frac\pi2=2\pi. \]
On the imaginary edge, traversed from \(iR\) down to \(0\), the image stays in the open right half-plane. Moreover \[ \arg f(iR)\longrightarrow0, \qquad \arg f(0)=0, \] so the change of argument along this edge tends to \(0\).
Choose \(R\) larger than the moduli of all roots. The number \(N_1\) of zeros in the first quadrant is then independent of \(R\), while the argument principle gives \[ 2\pi N_1=\Delta_{\Gamma_R}\arg f\longrightarrow2\pi. \] Therefore \(N_1=1\).
The coefficients of \(f\) are real, so nonreal roots occur in conjugate pairs. Hence there is also exactly one root in the fourth quadrant. Since \(f\) has degree \(4\) and no roots on the axes, the remaining two roots form a conjugate pair in the left half-plane, giving exactly one root in each of the second and third quadrants.