Past exam of the short options paper of the physics Prelims of the University of Oxford / 2019 / S1

A complex derivative computed along the real and imaginary directions gives $f^{\prime }=u_x+i{v}_x=v_y-i{u}_y$. Equality yields the Cauchy-Riemann equationsThe gradients satisfy $\nabla u\cdot \nabla v=u_x{v}_x+u_y{v}_y=0$. Since a gradient is normal to a regular level set, the two families meet at right angles by orthogonality wherever $f^{\prime }\ne 0$. At a critical point the regular-curve argument fails: for $f=z^2$ both gradients vanish at zero, so the assertion requires this qualification.

Set $A=x^2-y^2$ and $B=2xy$. Direct differentiation givesTheir sum vanishes, proving that $u$ is a harmonic function. More generally the Cauchy-Riemann equations give $u_{xx}+u_{yy}=v_{yx}-v_{xy}=0$ and $v_{xx}+v_{yy}=-u_{yx}+u_{xy}=0$; the mixed partial derivatives commute because a holomorphic function is smooth. Since $e^{i{z}^2}=e^{-B}(\cos A+i\sin A)$, a harmonic conjugate and the resulting holomorphic function areThese expressions also verify the Cauchy-Riemann equations directly.

The chain rule in polar coordinates gives $u_r=u_x\cos \theta +u_y\sin \theta$ and $u_{\theta }=r(-u_x\sin \theta +u_y\cos \theta )$, with analogous expressions for $v$. Substituting the Cauchy-Riemann equations givesThe restriction excludes the coordinate singularity at the origin.

Choose a continuous branch of a multivalued function of the complex logarithm on a simply connected region avoiding zero. ThenAlternatively $u_r=-e^{-\theta }\sin (\ln r)/r=v_{\theta }/r$ and $v_r=e^{-\theta }\cos (\ln r)/r=-u_{\theta }/r$, verifying the Cauchy-Riemann equations. The holomorphy is local to a chosen branch of a multivalued function: increasing $\theta$ by $2\pi$ multiplies $g$ by $e^{-2\pi }$, so this is not a single-valued holomorphic function on the whole punctured plane.

For $n\ge 1$, $f=z^n{e}^{n/z}$ is holomorphic at every finite nonzero point. At zero its Laurent series has infinitely many negative powers, so zero is an essential singularity. If the Riemann sphere is included, $f(1/w)=w^{-n}{e}^{nw}$ has a pole of order $n$ at $w=0$, so infinity is a pole of order $n$.

Expanding the exponential function produces the Laurent seriesIn particular its residue is $a_{-1}=n^{n+1}/(n+1)!$.

For a positively oriented circle of any radius $R>0$, the residue theorem givesOnly the $z^{-1}$ coefficient in the Laurent series contributes. Reversing the orientation changes the sign.

If $f(z)=\sum a_j{z}^j$, then $f(-z)=\sum (-1{)}^j{a}_j{z}^j$ and its residue at zero is $-a_{-1}$. The circle $|z-1|=2$ encloses zero and no other finite singularity. Thus the residue theorem givesfor positive orientation.

Cauchy's integral theorem states that a holomorphic function on a simply connected open region has zero contour integral around every closed piecewise smooth contour in that region. More generally the same holds for a contour homologous to zero in its domain of holomorphy. Having no singularity on the curve alone is insufficient: $1/z$ has a nonzero contour integral around the unit circle.

Put $z=e^{i\theta }$, so $d\theta =dz/(iz)$, $\cos 2\theta =(z^2+z^{-2})/2$ and ${\cos }^{2}3\theta =(z^6+2+z^{-6})/4$. The contour integral isThe enclosed poles are $0$ and $\pm 1/\sqrt{2}$. For the rational function displayed without the prefactor $1/(4i)$, the residue at zero is $-21/8$ and the two other residues sum to $27/8$. Therefore the residue theorem givesA useful independent check is the Fourier series $(5-4\cos \varphi {)}^{-1}=\frac{1}{3}[1+2{\sum }_{k\ge 1}{2}^{-k}\cos k\varphi ]$; only its constant and third Fourier coefficients survive.

The square root produces branch points at $z=\pm 1$, with local behaviour proportional to $(z\mp 1{)}^{-1/2}$. One branch cut can join them along $[-1,1]$. Another choice uses the two rays $(-\infty ,-1]\cup [1,\infty )$. Each choice prevents a loop from enclosing just one branch point within a single-valued domain. For the finite-segment choice fix $\sqrt{z^2-1}\sim z$ at infinity. Infinity is then a pole of order three, not a branch point, since

Use the branch cut $[-1,1]$ and the branch of a multivalued function normalized by $\sqrt{z^2-1}\sim z$. The Laurent series from the preceding solution has $z^{-1}$ coefficient $3/8$. Consequently every sufficiently large positively oriented circle givesAlthough $f$ grows cubically, its other Laurent series terms have zero contour integral. Choosing the opposite branch of a multivalued function reverses the sign; neither the branch of a multivalued function nor the orientation can be omitted from the answer.

Shrink the positive outer contour to the branch cut $[-1,1]$. On its upper bank the chosen branch of a multivalued function has $\sqrt{x^2-1}=i\sqrt{1-x^2}$ and on the lower bank it has the negative of this value. The upper bank runs from $1$ to $-1$, and the lower bank runs from $-1$ to $1$. The endpoint circles contribute $O(\sqrt{ϵ})$ and vanish. HenceThe change of variables $x=\sin t$ independently reduces the result to the integral of ${\sin }^{4}t$ over $[-\pi /2,\pi /2]$.