Show that the complex polynomial given by $z^4+2z^2-z+1$ has exactly one root in each quadrant.

I know by the fundamental theorem of Algebra, that the polynomial has exactly four roots.

Now, to show it has exactly one root in each quadrant, it suffices to show that there are two non-real roots in the open right and left half planes, since zeros of polynomials come in conjugate pairs, and they're just the reflection about the real axis in the complex plane.

I think I have to apply Rouché's theorem and compare $z^4+2z^2-z+1$ to a nice function, but I am not immediately sure what that function may be.

Any help would be much appreciated. Thanks in advance!


We usually apply Rouché's theorem inside the disk and here is not the case. Instead, we use the argument principle.

It suffices to show that there is exactly one root in the first quadrant because it is a polynomial with real coefficients, and zeros of polynomials come in conjugate pairs.

Denote $P(z)=z^4+2z^2-z+1$.

  • $P(z)$ has no zeros on the real axis because $P(z)=z^4+z^2+(z-\frac12)^2+\frac34>0$.

  • $P(z)$ has no zeros on the imaginary axis because $P(iy)=y^4-y^2-iy+1\ne0$.

  • Let $L_1=[0,R]$, $L_2=[0,Ri]$, $\Gamma=Re^{i\theta}, 0\le \theta\le\frac\pi2$. Consider the quarter circle $C=L_1+ \Gamma- L_2$. From the argument principle, $$\underbrace{\text{number of zeros}}_{:=N} - \underbrace{\text{number of poles}}_{=0}=\frac1{2\pi i}\oint_C\frac{P'(z)}{P(z)}~\mathrm dz=\frac{\Delta_C\arg P(z)}{2\pi},$$ or \begin{align*} N&=\frac{\Delta_{L_1}\arg P(z)}{2\pi}+\frac{\Delta_\Gamma\arg P(z)}{2\pi}-\frac{\Delta_{L_2}\arg P(z)}{2\pi}. \end{align*}

    We have $\Delta_{L_1}\arg P(z)=0$ because $P(x)>0~\forall~ x\in\Bbb R$, $$\Delta_\Gamma\arg P(z)=\Delta_\Gamma\arg z^4\left(1+\frac{2z^2-z+1}{z^4}\right)\to4\cdot\frac{\pi}{2}\quad\text{as}~ R\to\infty,$$ and \begin{align*} \Delta_{L_2}\arg P(z)&=\arg P(iR)=\arg{(R^4-R^2+1-iR)}\\ &=\arg\left(1-\frac{iR}{R^4-R^2+1}\right)\to 0 \quad\text{as}~ R\to\infty. \end{align*}

Therefore $N=1$, there is exactly one root in the first quadrant.

  • $\begingroup$ Thank you for the answer, it's much appreciated. What does the triangle (delta) mean though? $\endgroup$ – Math is Life Dec 6 '18 at 18:58
  • $\begingroup$ It is the variation of argument. The proof of Rouché's theorem is also based on this. $\endgroup$ – Tianlalu Dec 6 '18 at 19:28

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