T.3 · Complex analysis and residues

Undergraduate GRE preparation. Local objectives within the reviewed ETS scope; this is original teaching, not an official test or score predictor.

Prerequisites: Complex arithmetic, partial derivatives, series and oriented contours.

• Test complex differentiability
• Use contour integrals and residues
• Evaluate contour integrals with simple and higher-order pole residues

analytic 解析的: Complex differentiable throughout a neighbourhood.

residue 留数: Coefficient of (z−a)^−1 in a Laurent series.

Write f(z)=u(x,y)+iv(x,y). Complex differentiability imposes u_x=v_y and u_y=−v_x; with continuous first partials locally, the Cauchy–Riemann equations establish analyticity there. They differ from real differentiability of a two-coordinate map. The conjugate function x−iy fails these equations on every open neighbourhood, although real partial derivatives exist. State the region being checked, not only a convenient point.

An analytic function is complex differentiable throughout a neighbourhood and has a local convergent power series. A removable singularity can be filled analytically when the function is bounded near the missing point. A pole has a finite principal part in its Laurent series; an essential singularity has infinitely many negative-power terms. The residue is the coefficient of (z−a)⁻¹, not necessarily the leading or largest negative-power term.

For an isolated pole of order m, write f(z)=g(z)/(z−a)^m with g analytic at a. The residue is g^(m−1)(a)/(m−1)!. A simple pole uses g(a); a double pole uses g′(a). This follows by expanding g into its Taylor series. Thus e^z/(z−a)² has residue e^a, while a constant numerator over a pure double pole has zero residue. The pole order alone does not determine the contour integral.

For a positively oriented contour enclosing isolated singularities, the residue theorem gives ∮f(z)dz=2πi times the sum of enclosed residues, provided the function is analytic on the contour and elsewhere in the required interior. Reversing orientation changes the sign. Cauchy’s derivative formula is ∮g(z)/(z−a)^(m+1)dz=2πi g^(m)(a)/m! under its analytic-domain hypotheses. A singularity on the contour prevents direct application; a singularity outside contributes nothing to this contour.

For f(z)=conjugate(z), u=x and v=−y. Then u_x=1 but v_y=−1, so f is not complex differentiable. For 1/(z−2), a positively oriented circle |z−2|=1 encloses a simple pole of residue 1, hence the integral is 2πi.

Continuity or real differentiability alone does not imply complex analyticity.

Use the Cauchy–Riemann equations to compare $z^2$ and $\overline z$. Find the residue of $e^z/z^2$ at zero and its integral around the counterclockwise unit circle.

Worked solution

For $z^2$, $u=x^2-y^2$ and $v=2xy$: $u_x=2x=v_y$ and $u_y=-2y=-v_x$. The partials are continuous everywhere, so the function is analytic everywhere. For $\overline z$, $u=x,v=-y$ and $u_x=1\ne-1=v_y$; it is nowhere complex differentiable. Since $e^z=1+z+z^2/2+\cdots$, the coefficient of $z^{-1}$ in $e^z/z^2$ is 1. Therefore the residue is 1 and the integral is $2\pi i$. The double pole does not force a zero residue.

Integrate $1/[z(z-2)]$ around $|z|=1$, first counterclockwise and then clockwise. Can the same residue-theorem argument be used on $|z|=2$?

Check after attempting

Only zero is inside the unit circle. Its residue is $\lim_{z\to0}1/(z-2)=-1/2$. Thus the counterclockwise integral is $-\pi i$ and the clockwise integral is $\pi i$. The pole at 2 is outside and contributes nothing. The circle $|z|=2$ passes through a pole. The ordinary contour integral is not defined by this formula; the theorem requires analyticity on the contour. An indentation or principal-value prescription would be a different, explicitly specified problem.