Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2018 / S1 (Functions of a complex variable)

A branch point is a point around which analytic continuation changes the value of a multivalued function. For $z^{1/2}=r^{1/2}{e}^{i\theta /2}$, one circuit around zero changes the sign. A branch cut removes paths producing this ambiguity; for example $-\pi <\arg z<\pi$ gives a single holomorphic square root off the nonpositive real axis. The position of a cut is a convention, while the branch point is intrinsic.

Set $q(z)=(z-1)/(z+1)$. For $z$ outside the closed segment joining $-1$ to $1$, $q(z)$ avoids the nonpositive real axis. Therefore the principal square root of $q$ defines a holomorphic function there, with the normalization $f(\infty )=1$. The endpoints are branch points. Expanding at infinity,For a positively oriented contour surrounding the cut, its integral equals the integral on a sufficiently large positively oriented circle, since the intervening annulus is holomorphic. ThusOn the upper and lower banks, $f_{+}(x)=+i\sqrt{(1-x)/(1+x)}$ and $f_{-}(x)=-i\sqrt{(1-x)/(1+x)}$. The upper bank is traversed from $1$ to $-1$, and the lower bank from $-1$ to $1$. Their sum is $-2i$ times the required real integral. Small endpoint arcs vanish because their singularities have exponent less than one. ConsequentlyFor $\alpha =1/n$, use $f_{\alpha }=q^{\alpha }$ with the same normalization. Its expansion is $1-2\alpha /z+O(z^{-2})$, and its bank values are $e^{\pm i\pi \alpha }[(1-x)/(1+x){]}^{\alpha }$. Equating the contour integral $-4\pi i\alpha$ with $-2i\sin (\pi \alpha )$ times the real integral givesThe condition $n\ge 2$ ensures endpoint integrability. Choosing the other square-root branch reverses $I_{\Gamma }$, but also reverses both bank values, leaving the real integral unchanged.

Integrate $e^{iz}/z$ around an upper semicircle, indenting the origin by a small upper semicircle traversed clockwise. The integrand is holomorphic inside the indented contour. The large arc vanishes by Jordan's lemma, and the small arc contributes $-i\pi$, soTaking imaginary and real parts yieldsThe first integrand has a removable singularity at zero and its tails converge conditionally by Dirichlet test for improper integrals. The second integral needs a principal value at zero; its zero also follows from oddness with symmetric exclusion. One must not assign an ordinary convergent integral to its singularity.

Put $z=e^{i\theta }$, so $\sin \theta =(z-z^{-1})/(2i)$ and $d\theta =dz/(iz)$. The integral becomesThe poles are $-2i/3$ and $-3i/2$; only the first is inside the unit circle. The derivative of the denominator there is $12(-2i/3)+13i=5i$. The residue theorem therefore givesThere are no poles on the contour because the real denominator is at least one.

At a regular point, $f(z_0+\zeta )-f(z_0)=f^{\prime }(z_0)\zeta +O({\zeta }^2)$. Multiplication by a nonzero complex number rotates every tangent by the same angle and multiplies every length by the same positive factor. Oriented angles are preserved. If the first derivative vanishes but the second does not, the leading term is $f^{\prime \prime }(z_0){\zeta }^2/2$: tangent arguments are doubled, so angles are doubled, modulo the identification of rays. This is a local statement; holomorphy alone does not imply global injectivity.

Write $z=x+iy$. ThenFor $-1<x<1$ and $y>0$, the imaginary part is positive. The bottom edge maps monotonically to $[-1,1]$; the left and right vertical edges map to $(-\infty ,-1]$ and $[1,\infty )$ respectively. The inverse below establishes that every point of the open upper half-plane occurs exactly once. At fixed $y=a$, putting $u=\mathrm{Re}w$, $v=\mathrm{Im}w$, one obtainsThis is the upper half of an ellipse, traversed from its left endpoint to its right endpoint.

Choose the branch $s(w)=\sqrt{1-w^2}$ analytic in the upper half-plane and positive on the boundary interval $(-1,1)$. Then the inverse branch into the strip isThe logarithm is chosen continuously so that $-1<\mathrm{Re}z<1$ and $\mathrm{Im}z>0$; for instance $w=iv$, $v>0$, gives $z=(2i/\pi )\mathrm{arsinh}v$. Differentiation gives $dz/dw=2/(\pi \sqrt{1-w^2})$. The only zeros of the original derivative in the closed strip are $z=\pm 1$, with images $w=\pm 1$. The inverse has square-root branch points there, and cannot extend holomorphically across them. At either corner the original map doubles the right angle to a straight angle; a holomorphic inverse with nonzero derivative could not halve it. The open strip interior has no such failure.