Determine whether the equation \(z^3 + z^4 = 2\) in the complex variable \(z\) has any non-real solutions with \(|z| < 2\).
Solution
Yes. There are exactly two nonreal solutions, and both satisfy \(|z|<\sqrt2<2\).
1. Factoring off the root \(1\) leaves a cubic with exactly one real root.
The polynomial factors as \[ z^4+z^3-2=(z-1)q(z),\qquad q(z)=z^3+2z^2+2z+2. \] For real \(x\), \[ q'(x)=3x^2+4x+2 =3\left(x+\frac23\right)^2+\frac23>0. \] Hence \(q\) is strictly increasing on \(\mathbb R\). Since \[ q(-2)=-2<0<1=q(-1), \] the intermediate value theorem gives a root \(r\in(-2,-1)\), and strict monotonicity makes it the unique real root of \(q\). Because \(q\) has real coefficients, its remaining two roots are a nonreal conjugate pair \(\alpha,\overline\alpha\).
2. Vieta’s formula places the conjugate pair strictly inside the radius-two disk.
For the monic cubic \(q\), the product of its three roots is \(-2\). Thus \[ r\alpha\overline\alpha=-2, \qquad |\alpha|^2=-\frac2r. \] Since \(-2<r<-1\), \[ 1<-\frac2r<2. \] Therefore \[ |\alpha|=|\overline\alpha|<\sqrt2<2. \] Thus the original quartic has exactly two nonreal roots and both lie in the specified disk.