Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 1 / 11G (Topics in Analysis)

Parametrize the circle by $z=e^{it}$, $0\le t\le 2\pi$. Choose a continuous argument $\vartheta (t)$ of $\varphi (e^{it})$, so $\varphi (e^{it})=|\varphi (e^{it})|e^{i\vartheta (t)}$. Its winding number about zero isThe two facts proved below determine the requested calculation without needing differentiability or contour integration. Put $P(z)=(3z-2)(z-3)(2z+1)$. On $|z|=1$, $|P(z)|\ge 1\cdot 2\cdot 1=2$, so adding one does not change its winding. The first and third factors have winding one, since their constant terms have modulus smaller than their linear terms; the second has winding zero, since its $z$ term has modulus smaller than its constant term. Nonzero constants have winding zero and $z$ has winding one. Thus

Choose continuous arguments ${\vartheta }_1,{\vartheta }_2$ along the parametrized circle. Their sum is a continuous argument of the product, because multiplication multiplies moduli and adds arguments. Subtracting endpoint values gives

Factor $\varphi +\psi =\varphi (1+\psi /\varphi )$. The second factor has positive real part, since $|\psi /\varphi |<1$. It therefore has the continuous single-valued argument $\arctan (\Im (1+\psi /\varphi )/\Re (1+\psi /\varphi ))$ taking values in $(-\pi /2,\pi /2)$. Its values agree at the two ends of the parametrized circle, so its winding number is zero. Part (i) gives