Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2013 / Analysis / 6

An isolated singularity at $a$ means that $f$ is holomorphic in some punctured disk $0<|z-a|<r$, while $a$ itself is an excluded point or a point where the given function is not holomorphic. There are no other singularities arbitrarily close to $a$. By Laurent's theorem, the function has a unique Laurent series on that punctured disk:The negative-power terms form its principal part of a Laurent series.

A removable singularity is an isolated singularity for which the function on the punctured disk extends to a holomorphic function on the full disk, by assigning or correcting its value at $a$. In Laurent series terms,The remaining power series defines the extension with value $c_0$ at $a$. Conversely, a holomorphic extension has a Taylor series there, so uniqueness of the Laurent series forces all negative coefficients to vanish.

A pole of order $m\ge 1$ at $a$ has finitely many negative powers in its Laurent series, the lowest being exactly $-m$:Equivalently,Thus $(z-a{)}^{m}f(z)$ has a holomorphic extension with nonzero value at $a$, whereas multiplying by any smaller power does not remove the singularity.

A zero of order $n\ge 1$ at $a$ means that $f$ is holomorphic near $a$ andEquivalently its Taylor series starts with the nonzero coefficient of $(z-a{)}^n$, soUnlike a pole, a zero is a regular point of the function itself; its reciprocal has a pole of the corresponding order.

By the finite-order zero factorizations,with $F,G$ holomorphic near $a$. Shrinking the disk if necessary, continuity ensures $G$ never vanishes there, so $F/G$ is holomorphic and nonzero at $a$. ConsequentlyIf $n>m$, the exponent is $-(n-m)$, with a holomorphic nonvanishing factor. Hence

With the same factors, if $n\le m$, thenhas no negative power and extends holomorphically across $a$. Thus its apparent singularity is a removable singularity, with extension valueFor $m>n$ the extended quotient has a zero of order $m-n$; for $m=n$ the extension is nonzero.

The complex sine has simple zeros exactly at $z=k\pi$, $k\in \mathbb{Z}$, because $\sin (k\pi )=0$ and $\cos (k\pi )=(-1{)}^k\ne 0$. Hence the denominator has a zero of order three at each of these points and nowhere else.

At $0$, the numerator has order two, so the quotient has a simple pole. Indeed its leading term is $-{\pi }^4/z$. At $\pi$, the numerator has order one, so there is a pole of order two. At $-\pi$, the numerator has order three, cancelling the denominator's order-three zero, so the singularity is removable. Its extension value isAt every remaining $k\pi$, the numerator is nonzero, giving a pole of order three. Thus the complete finite-plane classification isAway from this discrete set the denominator is nonzero, so the quotient is holomorphic.