Prove that \(z^4 + 2 z^3 - 2z + 10 =0\) has exactly one root in each open quadrant.
Solution
Let \(P(z) = z^4 + 2z^3 - 2z + 10\). For \(R>0\), let \(Q_R=\{z:\abs{z}<R,\ 0<\arg z<\pi/2\}\), whose boundary consists of \(\gamma_1=[0,R]\), the arc \(\gamma_2\): \(z=Re^{i\theta}\), \(0\le\theta\le\pi/2\), and the segment \(\gamma_3\) from \(iR\) to \(0\).
1. \(P\) has no real root.
For real \(x\), \(P(x) = (x^2 + x - 1)^2 + x^2 + 9 \ge 9\).
2. \(P\) has no root on the imaginary axis, and \(\operatorname{Re}P(iy)>0\) for all real \(y\).
\(P(iy) = (y^4 + 10) - 2iy(y^2 + 1)\), whose real part is at least \(10\).
3. For \(R\ge3\) and \(\abs{z}=R\), \(\abs{P(z)-z^4} < \abs{z^4}\). In particular every root of \(P\) satisfies \(\abs{z}<3\).
\(\abs{P(z) - z^4} \le 2R^3 + 2R + 10\), and \(R^4 - 2R^3 - 2R - 10\) is positive at \(R=3\) (it equals \(11\)) and increasing for \(R\ge3\), since its derivative \(4R^3-6R^2-2\) is positive there.
4. For \(R\ge3\), a continuous argument of \(P\) increases by exactly \(2\pi\) around \(\partial Q_R\).
4.1. Along \(\gamma_1\) the argument does not change.
\(P>0\) on \([0,R]\) by step 1 (and \(P(0)=10\)).
4.2. Along \(\gamma_2\) the argument increases by \(2\pi+\delta\), where \(\delta\in(-\pi/2,\pi/2)\) is the principal argument of \(P(iR)\).
Write \(P(z) = z^4\,w(z)\) with \(w(z) = P(z)/z^4\). By step 3, \(\abs{w(z)-1}<1\) on \(\gamma_2\), so \(w\) stays in the right half-plane and its principal argument is a continuous argument there. It is \(0\) at \(z=R\) (where \(w>0\)) and \(\delta\) at \(z=iR\) (where \(w=P(iR)/R^4\)). The factor \(z^4\) contributes \(4\cdot\pi/2=2\pi\).
4.3. Along \(\gamma_3\) the argument changes by \(-\delta\).
By step 2, \(P(iy)\) lies in the right half-plane for every \(y\), so the principal argument is continuous along \(\gamma_3\). It equals \(\delta\) at \(iR\) and \(0\) at \(P(0)=10\).
4.4. Q.E.D.
Add steps <2>1–<2>3: \(0 + (2\pi+\delta) - \delta = 2\pi\).
5. \(P\) has exactly one root in the open first quadrant.
\(P\) has no zeros on \(\partial Q_R\) by steps <1>1–<1>3. By the argument principle and step 4, \(P\) has exactly one zero in \(Q_R\) for every \(R\ge3\), and by step 3 all zeros lie in \(\abs{z}<3\).
6. \(P\) has exactly one root in each open quadrant.
\(P\) has real coefficients, so \(z\mapsto\bar z\) maps its roots in the first quadrant bijectively onto its roots in the fourth; by step 5 there is exactly one in each. \(P\) has degree \(4\) and no roots on the axes (steps 1 and <1>2), so the remaining two roots lie in the open second and third quadrants, and conjugation again matches them, giving one in each.
7. Q.E.D.
Step 6 is the required statement.