Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2008 / Paper 1 / 7

For $f=u+iv$ holomorphic in $z=x+iy$, the Cauchy-Riemann equations are $u_x=v_y$ and $u_y=-v_x$. Therefore the normal vectors to the two level curves obeyAt a point where $f^{\prime }\ne 0$, both gradients are nonzero, so their perpendicular tangent lines are also perpendicular. At critical points this argument does not assert that regular level curves exist.

TakeIts real part is $u=x^2-y^2$, so the specified first family is $u=-\alpha$. Its imaginary part is $v=2xy$, so the second family is $v=2\beta$. Since $\beta \ne 0$, their intersections exclude the origin, where $f^{\prime }(z)=2z$ would vanish. Thus these hyperbolas meet orthogonally as required.

For an isolated pole of order $N\ge 1$, the Laurent coefficients must satisfyThen the principal part of a Laurent series is a finite sum with lowest power $-N$. The classification at the center assumes analyticity on a punctured neighborhood $0<|z-z_0|<b$, or an analytic continuation there. A Laurent expansion on an annulus with positive inner radius alone does not imply that the center is an isolated singularity.

On a punctured neighborhood of the center, an essential singularity occurs exactly when infinitely many negative-power Laurent coefficients are nonzero. If all negative-power coefficients vanish, the singularity is removable; if finitely many are nonzero, the singularity is a pole. The punctured-neighborhood assumption matters: an annulus separated from its center cannot by itself classify the function there.

Put $w=z-i\pi$. Since $\sinh (w+i\pi )=-\sinh w$ and $\sinh w=w(1+w^2/6+O(w^4))$, the local Laurent series isThus all singular terms areThere is a third-order pole, with no second-order term.

Use the local coordinate $w=1/z$ at infinity. For positive integer $N$,The negative powers of this local coordinate are precisely $m=0,\dots ,N-1$. The singular part at infinity is thereforeInfinity is a pole at infinity of order $N$. In the original $z$ variable these singular terms are positive powers; it is the negative powers of $w=1/z$ that define the local principal part of a Laurent series. For $N=0$ the point at infinity is regular, and for negative integer $N$ it is a zero rather than a pole.

The exponential power series gives $e^{-1/z}={\sum }_{m=0}^{\infty }(-1{)}^m/(m!z^m)$ for $z\ne 0$. Hence all singular terms areInfinitely many negative powers occur, so zero is an essential singularity.