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Let $N(t)$ count roots in the open unit disk. A boundary crossing can occur only at the real values $0,\pm 1,\pm 2$ found above, so the count is constant between them. Also $N(-t)=N(t)$, because the polynomial is odd. At $t=0$, $p(z)=z(z^4+1)$ has one root inside and four on the circle, giving $N(0)=1$.

For small positive $t$, the root near zero stays inside. At each boundary root ${\zeta }^4=-1$, $p^{\prime }(\zeta )=-4$, so implicit differentiation gives $z(t)=\zeta -t/4+O(t^2)$. The radial crossing test for polynomial root counts isThe two roots with positive real part enter and the other two leave. Hence $N(t)=3$ for $0<t<1$.

At $t=1$, the boundary roots are $\zeta =e^{\pm i\pi /3}$. Since $\zeta p^{\prime }(\zeta )=5t-4\zeta$, their radial derivative is $\mathrm{Re}(1/(5-4\zeta ))=1/7>0$. Both leave as $t$ increases, so the count becomes one; at $t=1$ itself they are excluded, also leaving one interior root. At $t=2$, the remaining boundary root $z=1$ has radial derivative $1/6>0$ and leaves. For $t\ge 2$ no interior root is possible, since $|p(z)|\le |z{|}^5+|z|<2$ in the open disk. Combining these facts and odd symmetry givesThis also gives the corresponding winding numbers of the image contour away from its crossings. Boundary roots are excluded at the crossing values; using a nearby winding number at $t=0$ would incorrectly count two extra roots.