Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 4 / 2G / Solution

For $t<1/2$, the number $f(t)$ is negative real, so the proposed identity would forceContinuity on the connected interval $[0,1/2)$ makes $\varphi$ equal to one fixed odd multiple of $\pi$ there. For $t>1/2$, the number $f(t)$ is positive real, so $\varphi$ must instead equal one fixed even multiple of $\pi$. These two constants cannot agree, and their one-sided limits at $1/2$ therefore contradict continuity. The fact that $f(1/2)=0$ removes the phase condition at that one point but does not repair the discontinuity.

Now let $g:[0,1]\to \mathbb{C}\setminus \{0\}$ and put $h=g/|g|$. This is a continuous path in the unit circle. The exponential map $\theta \mapsto e^{i\theta }$ is a covering map, so the path lifting theorem supplies a continuous lift $\theta :[0,1]\to \mathbb{R}$ after one value of $\theta (0)$ is chosen. HenceOne can obtain the same lift directly from the allowed special case: by uniform continuity, subdivide $[0,1]$ so that on each subinterval $h(t)/h(t_j)$ lies in the right half-plane, choose its continuous local argument there, and add a multiple of $2\pi$ to match the preceding endpoint.

If $\theta$ and $\tilde{\theta }$ are two such phases, thenThis integer-valued function is continuous and hence constant. Thereforedoes not depend on the chosen lift.

For $u(t)=g(t^2)$, the phase $\theta (t^2)$ givesFor $v(t)=g(t{)}^2$, the phase $2\theta (t)$ gives

Finally, takeThey have the same two endpoints, but $r(g_1)=0$ and $r(g_2)=2\pi$. This endpoint discrepancy records the winding number of the closed path.