Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2011 / Analysis

For the preliminary definition, put $S_N(z)={\sum }_{k=1}^N{f}_k(z)$. The series converges uniformly to $f$ on $U$ when, for every $ϵ>0$, some $N_0$ satisfies $|S_N(z)-f(z)|<ϵ$ for all $z\in U$ and all $N\ge N_0$. Equivalently, ${\sup }_{z\in U}|S_N(z)-f(z)|\to 0$.

Let $L(\gamma )$ be the finite length of the piecewise-smooth path. Finite linearity of the contour integral and the integral bound giveThe uniform limit is continuous by the stated allowed result, so its integral is defined. Consequently

Cauchy's integral theorem states that if a function is holomorphic on a simply connected open set, its integral around every closed piecewise-smooth path in that set is zero. A local form sufficient here is: if a function is holomorphic on an open set containing a triangle together with its interior, thenFor a general domain with holes, an arbitrary closed path need not have zero integral; the triangle or simple-connectedness hypothesis supplies the required interior.

Morera's theorem states that if $f$ is continuous on an open set $U$ and its integral around every triangle whose interior and boundary lie in $U$ is zero, then $f$ is holomorphic on $U$.

For a proof, choose a disk $B\subseteq U$ and a base point $z_0\in B$. Define $F(z)={\int }_{[z_0,z]}f(\zeta )d\zeta$ along the straight segment. Convexity keeps every triangle with vertices $z_0,z,z+h$ inside $B$. Its zero boundary integral impliesBy continuity, the last integral tends to $f(z)$ as $h\to 0$, so $F^{\prime }(z)=f(z)$. Thus $F$ is holomorphic on $B$. The allowed result that the derivative of a holomorphic function is holomorphic now gives that $f=F^{\prime }$ is holomorphic there. Such disks cover $U$, proving

For each triangle compactly contained in $U$, part (i) permits the interchange of the series and its boundary integral. Cauchy's integral theorem makes every term zero. The uniform limit is continuous, so Morera's theorem yields that $f$ is holomorphic on $U$.

For the final application, each function $(z-k{)}^{-2}$ is holomorphic away from its pole. Fix a positive integer $n$. For $z\in U_n$ and $k>2n$,The numerical series ${\sum }_{k>2n}4/k^2$ converges, so the Weierstrass M-test gives uniform convergence of the tail on $U_n$. Adding the finitely many initial terms preserves uniform convergence. Those initial terms need not be bounded near their excluded poles; only the tail needs the uniform bound. By the first conclusion, the full sum is holomorphic on each $U_n$. Their union is $\mathbb{C}\setminus \mathbb{N}$, hence

The residue at $a$ is the coefficient $c_{-1}$ of $(z-a{)}^{-1}$ in the Laurent series. A simple pole means $f(z)=c_{-1}/(z-a)+h(z)$ near $a$, with $c_{-1}\ne 0$ and $h$ holomorphic across $a$.

For the specified positively oriented arc, direct parametrization gives ${\int }_{{\gamma }_{ϵ}}(z-a{)}^{-1}dz=i(\beta -\alpha )$. The remaining integral is bounded by $ϵ(\beta -\alpha )\sup |h|$, which tends to zero because $h$ is bounded near $a$. ThereforeThe simple-pole hypothesis matters. For $f(z)=(z-a{)}^{-2}$, with $\alpha =0$, $\beta =\pi$, the residue is zero butFor the particular function, expanding the exponential givesThus $\boxed{\mathrm{res}(f;0)=1/2}$, and the pole is simple. On the large upper semicircle, $|e^{iz}|\le 1$, givingClose the real-axis path from $-R$ to $-ϵ$ and $ϵ$ to $R$ by this upper arc and a clockwise upper indentation at zero. The function is holomorphic in the indented region, so the sum of all four integrals is zero. The clockwise indentation tends to $-i\pi /2$. For real $x$,The real part is odd and cancels on the symmetric real pieces; the imaginary part is even. Its integral converges since $(x-\sin x)/x^3\to 1/6$ at zero and is $O(x^{-2})$ at infinity. Letting $ϵ\to 0$ and $R\to \infty$ therefore yields $2iI-i\pi /2=0$, where $I$ is the requested real integral. Hence

A conformal map preserves angles between intersecting regular curves, including their orientation. If $f$ is holomorphic with $f^{\prime }(a)\ne 0$, the image tangent vectors are $f^{\prime }(a)$ times the original tangent vectors. Multiplication by this nonzero complex number gives a common positive scaling and rotation, so their oriented angle is unchanged. This proves conformality throughout the stated domain.

At a zero derivative this can fail: $f(z)=z^2$ at zero sends rays of arguments $0$ and $\pi /4$ to rays of arguments $0$ and $\pi /2$. The angle is doubled, so this holomorphic function is not conformal there.

For the disk map $C(z)=i(1+z)/(1-z)$, computeIts inverse is $z=(w-i)/(w+i)$; for $\mathrm{Im}w>0$, the inequality $|w-i|<|w+i|$ puts this inverse in the unit disk. Both rational maps are holomorphic on their respective domains. ThusFor the final region, writing $z=x+iy$ gives $|e^z|=e^x\in (0,1)$ and $\arg (e^z)=y\in (0,\pi )$. Hence its exponential image is exactly the upper half of the unit disk,The exponential is injective on this strip, since two equal exponential values differ by $2\pi i$ times an integer while the imaginary parts here differ by less than $\pi$. Its derivative never vanishes and its inverse on $V$ is the logarithm with argument in $(0,\pi )$.

Now $L(w)=(1+w)/(1-w)$ maps $V$ conformally onto the first quadrant. Indeedand the inverse $(s-1)/(s+1)$ has modulus below one and positive imaginary part for every first-quadrant $s$. Squaring maps that quadrant bijectively onto $\mathbb{H}$, with nonzero derivative. Therefore a required composition isEvery component is a conformal bijection between the stated domains, so the composition is one as well.