Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2010 / Analysis

Write $f=u+iv$ and $z=x+iy$. The complex derivative computed along a real increment is $f^{\prime }=u_x+i{v}_x$, while along an imaginary increment it is $f^{\prime }=v_y-i{u}_y$. Equality givesThese are the Cauchy–Riemann equations at every point where $f$ is holomorphic. The directional calculations also establish the existence of those partial derivatives. More explicitly, $f(z+h)-f(z)=f^{\prime }(z)h+o(|h|)$ gives both real-direction formulas directly. If a converse is desired, continuous first partial derivatives together with these equations make the real differential multiplication by $u_x+i{v}_x$, hence give complex differentiability; the necessity above is what is requested.

The allowed theorem makes $f^{\prime }$ holomorphic and hence continuous; applying it repeatedly gives continuous derivatives of every order. Since $u_x=\Re f^{\prime }$, $v_x=\Im f^{\prime }$, $u_y=-\Im f^{\prime }$ and $v_y=\Re f^{\prime }$, the real and imaginary parts have continuous second partial derivatives. Differentiate the Cauchy–Riemann equations and commute mixed derivatives:Thus $\boxed{\Delta u=\Delta v=0}$, with the required $C^2$ regularity, proving harmonicity.

Write $f=u+iv$ and evaluate all derivatives of $\psi$ at $(u,v)$. The real chain rule givesThe Cauchy–Riemann equations give equal squared lengths $u_x^2+u_y^2=v_x^2+v_y^2=|f^{\prime }{|}^2$ and zero cross term $u_x{v}_x+u_y{v}_y=0$. Part (ii) removes the last two terms. ThereforeThe composition is $C^2$, so it is harmonic. No assumption that $f^{\prime }$ is nonzero is required.

The complex exponential function $E(z)=e^z$ maps the strip bijectively and holomorphically to the upper half-plane: its modulus is $e^{\Re z}>0$ and its argument is $\Im z\in (0,\pi )$. Its inverse is the holomorphic logarithm $L(w)=\log |w|+i\mathrm{Arg}w$ with $\mathrm{Arg}w\in (0,\pi )$.

Part (iii) applies to both maps. The required inverse correspondences areThey compose to the identity because $L\circ E$ and $E\circ L$ do. This proves injectivity and surjectivity, not merely a map between the two sets.

For any $0<r<1$, a Laurent series converges uniformly on the circle $|z|=r$, so termwise contour integration is valid. Multiplication by $z^{-k-1}$ and integration gives, for every integer $k$,Indeed, the integral of $z^{j-k-1}$ is $2\pi i$ for $j=k$ and zero otherwise. These coefficients depend only on $f$, so any two Laurent expansions must agree coefficient by coefficient. This proves uniqueness, including all terms of the principal part.

A removable singularity means that $f$ has a holomorphic extension to the full disk, assigning a finite value at zero. Equivalently its Laurent series has no negative-power terms. One direction follows by the Taylor expansion of the extension; the other by the convergent nonnegative-power series. Thus $\boxed{a_k=0\text{ for all }k<0}$ is the coefficient criterion.

A pole of order $n$ means $f(z)=z^{-n}g(z)$ with $g$ holomorphic near zero and $g(0)\ne 0$. Equivalently, the principal part is finite and starts with $\boxed{a_{-n}{z}^{-n},a_{-n}\ne 0}$, while $a_k=0$ for $k<-n$. In particular $n$ is the exact order, not merely an upper bound.

An essential singularity is an isolated singularity that is neither removable nor a pole. In terms of the Laurent series this means $\boxed{\text{infinitely many negative-power coefficients are nonzero}}$: no finite multiplication by a power of $z$ removes the singularity.

For $|z|<1$, differentiate the geometric series to obtain $(1-t{)}^{-2}={\sum }_{k\ge 0}(k+1)t^k$. Since $(z-i{)}^2=-(1-z/i{)}^2$,The convergence region is precisely $0<|z|<1$, the specified punctured disk; the other pole is on its boundary at $i$.

If $f=z^{-n}g$ with $g(0)\ne 0$, continuity gives a small disk on which $|g(0)|/2\le |g(z)|\le 3|g(0)|/2$. Multiplying by $|z{|}^{-n}$ gives the two required bounds.

Conversely, set $g(z)=z^{n}f(z)$ on the punctured disk. The upper bound makes $g$ bounded, so the allowed removable-singularity theorem gives a holomorphic extension. The lower bound gives $|g(0)|\ge C_1>0$ by continuity. Therefore $f=z^{-n}g$ has a pole of exact order $n$. The lower bound is essential: the upper bound alone only proves order at most $n$.

On $D=\mathbb{C}\setminus [0,\infty )$ choose the continuous argument $0<\arg z<2\pi$, and putThis is holomorphic on the slit plane for every fixed $w$. The upper boundary value at $x>0$ is $x^w$, whereas the lower boundary value is $e^{2\pi iw}{x}^w$. This difference is exactly what the keyhole contour measures.

Use the branch in (i) and the positively oriented slit-annulus contour. Its upper ray runs from $r$ to $R$, its outer circle counterclockwise, its lower ray from $R$ back to $r$, and its inner circle clockwise. Initially keep their arguments at $\alpha$ and $2\pi -\alpha$ as in the figure, with $0<\alpha <\pi /2$; let $\alpha \downarrow 0$ for fixed $0<r<1<R$. Both poles $i,-i$ lie inside.

Write $\sigma =\Re w\in (-1,1)$. On the outer circle the absolute integral is $O(R^{\sigma -1})$, since $|z^w|\le e^{2\pi |\Im w|}{R}^{\sigma }$ and $|z^2+1|\ge R^2-1$. On the inner circle it is $O(r^{\sigma +1})$. Both tend to zero. The real integral converges absolutely because its endpoint behaviors are $x^{\sigma }$ and $x^{\sigma -2}$.

The two residues, with the chosen arguments, areThe residue theorem and the two ray orientations therefore giveFor $w\ne 0$ in this strip we may cancel $1-e^{i\pi w}$; its only possible zero here is $w=0$. ConsequentlyAt $w=0$ the same result follows directly from $I(0)={\int }_0^{\infty }(1+x^2{)}^{-1}dx=\pi /2$, or by dominated convergence. The denominator has no zero in the open strip.

Set $w=i$, which satisfies $\Re w=0$, and take real parts of the absolutely convergent integral. Since $x^i=\cos (\log x)+i\sin (\log x)$ and $\cos (i\pi /2)=\cosh (\pi /2)$,Absolute integrability justifies moving the real-part operation through the integral.