Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2006 / Analysis

A function is holomorphic on an open set if it is complex differentiable at every point of that set. At $z_0=x_0+i{y}_0$, take difference quotients first along real increments. Their limit is $f^{\prime }(z_0)$, so $u_x=\mathrm{Re}{f}^{\prime }(z_0)$ and $v_x=\mathrm{Im}{f}^{\prime }(z_0)$. Along imaginary increments,so $u_y=-\mathrm{Im}{f}^{\prime }(z_0)$ and $v_y=\mathrm{Re}{f}^{\prime }(z_0)$. These limits prove existence of all four partial derivatives and the Cauchy-Riemann equations $u_x=v_y$, $u_y=-v_x$.

For the first example, the identity function and the reciprocal function are holomorphic away from zero, so $f$ is holomorphic on the given domain, with

Write $r=\sqrt{x^2+y^2}>0$. For $u=x/(1+r)$, $v=y/(1+r)$,The two Cauchy-Riemann equations would require $x^2=y^2$ and $xy=0$ simultaneously, impossible when $r>0$. Thus $f$ is not complex differentiable at any nonzero point.

At zero, however,So $f$ is complex differentiable at zero, with derivative one, but is not holomorphic on $\mathbb{C}$ or on any nonempty open subset. Complex differentiability at a single point is weaker than holomorphy in a neighborhood.

Choose a small positively oriented circle $C_{\epsilon }$ about $z_0$ inside the given region. Cauchy's theorem on the region between the two contours givesContinuity makes the last integrand converge uniformly to $f(z_0)$ as $\epsilon \downarrow 0$. Its limit is $2\pi if(z_0)$, proving the Cauchy integral formula.

The same formula holds with an interior point $w$ near $z_0$. Subtract its value at $z_0$ and divide by $w-z_0$:The contour is a positive distance from $z_0$, so the integrands converge uniformly as $w\to z_0$. Passing to the limit gives the differentiated formulaOn the unit circle, $\bar{z}=1/z$ and $\mathrm{Im}z=(z-\bar{z})/(2i)$, proving the requested identity. Hence the requested integral isThe first integral is $2\pi i/3$. For the second useThe differentiated formula applied to the constant function gives $\int z^{-2}dz=0$; the remaining two terms integrate to $-6\pi i+6\pi i=0$. Therefore

At a zero $a$, Taylor's theorem gives $f(z)={\sum }_{n\ge 0}{c}_n(z-a{)}^n$ near $a$. If some coefficient is nonzero, let $m$ be the least such index. Then $f(z)=(z-a{)}^{m}q(z)$ with $q(a)\ne 0$, and continuity makes $q$ nonzero on a smaller disc. Thus $a$ is an isolated zero. A limit point of zeros is also a zero by continuity and cannot be isolated, so all Taylor coefficients there vanish and $f$ vanishes on a nonempty open disc.

Let $A$ be the set of points having a neighborhood on which $f$ vanishes. It is nonempty and open. A point in its relative closure either already belongs to $A$ or is a limit point of distinct zeros; the preceding argument puts it in $A$ in either case. Thus $A$ is also relatively closed. Connectedness of $D$ gives $A=D$, proving $f$ is identically zero.

For conjugate reflection, the disc is invariant under conjugation, andHence $h$ is holomorphic, with derivative $\overline{g^{\prime }(\bar{z})}$. If $g$ is real on the real diameter, then for real $x\in (-1,1)$, $h(x)=\overline{g(x)}=g(x)$. The holomorphic difference $g-h$ therefore has zeros accumulating at zero, and the identity theorem just proved gives $g=h$ on the entire disc.