Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2004 / Analysis

Writing $z=x+iy$ and $f=u+iv$, the Cauchy-Riemann equations areWith continuous second derivatives, mixed derivatives commute. ThusBoth real and imaginary parts are harmonic.

For $g(z)=\mathrm{Re}z$, the real and imaginary parts are $x$ and zero, so $u_x=1\ne 0=v_y$. For $h(z)=\bar{z}$, they are $x$ and $-y$, so $u_x=1\ne -1=v_y$. The necessary Cauchy-Riemann equations fail at every point. Hence both functions are nowhere holomorphic, indeed nowhere complex differentiable.

A removable singularity at $a$ means that there is a holomorphic function on the whole disc which agrees with $f$ away from $a$. Equivalently, the Laurent series has no negative powers. The value at $a$ is filled in by its finite limit.

A pole of order $m\ge 1$ means that $(z-a{)}^{m}f(z)$ extends holomorphically to $a$ and has a nonzero value there. Equivalently, the lowest nonzero power in the Laurent expansion is $(z-a{)}^{-m}$.

An essential singularity is an isolated singularity which is neither removable nor a pole; equivalently its Laurent expansion has infinitely many nonzero negative-power terms.

For the final unnumbered classification, the only possible finite singularities are the zeros of the denominator, namely $z=\pm i$ and $z=k\pi$ for integers $k$. At zero,so zero is removable, with extension value one. At each nonzero $k\pi$, $\sin z$ has a simple zero, the other denominator factor is nonzero and the numerator is nonzero. These points are poles of order two, with leading coefficient $(k\pi {)}^2/((k\pi {)}^2+1)$.

At $z=\pm i$, the polynomial denominator has a simple zero. Since $\sin (\pm i)=\pm i\sinh 1\ne 0$ and the numerator is $-1$, these are simple poles. There are no essential singularities on $\mathbb{C}$ and no other finite singularities.

After division by $z(z+2)$, the coefficients areTheir only finite singularities are $z=0$ and $z=-2$. At either point $z_0$, the products $(z-z_0)p$ and $(z-z_0{)}^{2}q$ are analytic, so both points are regular singular points. All other finite points are ordinary.

The lowest power after substituting the Frobenius series gives $\sigma (2\sigma -1)=0$. The coefficient of $z^{n+\sigma }$ then givesFor both exponents the denominator is nonzero for every $n\ge 0$, so each arbitrary nonzero $a_0$ determines a series. Unless it terminates, its coefficient ratio tends to $-1/2$, proving convergence for $|z|<2$. For the half-integer exponent choose a square-root branch. Their leading powers $1$ and $z^{1/2}$ show that the two local solutions are linearly independent.

If $\alpha =m$ is a positive integer, choose $\sigma =0$. The recurrence numerator is nonzero for $0\le n<m$ and vanishes at $n=m$. Thus $a_m\ne 0$, $a_{m+1}=0$, and all later coefficients vanish: a polynomial solution of exact degree $m$ exists. It is proportional to the Chebyshev polynomial $T_m(z+1)$, since the substitution $t=z+1$ transforms the equation into $(t^2-1)y^{\prime \prime }+t{y}^{\prime }-m^{2}y=0$.