Analytic Functions and Residues

Concise answer

Complex analysis on this exam is a short pipeline: check analyticity with the Cauchy-Riemann equations, locate the isolated singularities, keep only the ones inside the contour, compute each residue with the formula matching the pole order, and multiply the sum by $2\pi i$ . Cauchy's theorem is the special case where nothing is enclosed and the integral is zero.

Definitions

Analytic (holomorphic): Complex-differentiable at every point of an open set. Analytic functions are infinitely differentiable and equal to their Taylor series there.
Cauchy-Riemann equations: For $f = u + iv$ , the conditions $u_x = v_y$ and $u_y = -v_x$ . They are necessary for analyticity, and sufficient when $u$ and $v$ have continuous partial derivatives.
Pole of order m: An isolated singularity at $a$ whose Laurent series has a lowest term $(z-a)^{-m}$ with $m$ finite; $m = 1$ is a simple pole.
Essential singularity: An isolated singularity whose Laurent series has infinitely many negative-power terms, as $e^{1/z}$ has at $z = 0$ .
Residue: The coefficient of $(z-a)^{-1}$ in the Laurent expansion of $f$ about $a$ ; it is the only coefficient that survives integration around a small circle.
Entire function: A function analytic on the whole complex plane, such as $e^z$ , $\sin z$ , and every polynomial.

Intuition

The residue is the only Laurent coefficient with anywhere to go. Integrating $(z-a)^{n}$ around a circle centred at $a$ gives zero for every integer $n$ except $n = -1$ , where the answer is $2\pi i$ . So integrating a Laurent series term by term annihilates everything but one coefficient, and the residue theorem is that observation applied to each enclosed singularity at once.

Cauchy's theorem and the residue theorem are the same statement. If nothing singular is enclosed, the sum of residues is empty and the integral is zero; each enclosed singularity contributes exactly $2\pi i$ times its residue. That is why the very first question about any contour integral is which singularities lie inside — not how to integrate.

Concept walkthrough

Analyticity is a strong condition: complex differentiability on an open set forces infinite differentiability and a convergent power series. The working test is the Cauchy-Riemann pair $u_x = v_y$ , $u_y = -v_x$ for $f = u + iv$ . These are necessary for analyticity everywhere, and when $u$ and $v$ have continuous partials they are also sufficient. The standard non-example is $f(z) = \bar{z} = x - iy$ , where $u_x = 1$ but $v_y = -1$ , so $f$ is nowhere analytic even though it is perfectly smooth as a map of $\mathbb{R}^2$ .

Once analyticity is settled, integration becomes bookkeeping. Cauchy's theorem says a function analytic on and inside a simple closed contour integrates to zero around it. Cauchy's integral formula upgrades that to $f(a) = \dfrac{1}{2\pi i}\oint_C \dfrac{f(z)}{z-a}\,dz$ , and the derivative version handles higher powers in the denominator: $\oint_C \dfrac{f(z)}{(z-a)^{n+1}}\,dz = \dfrac{2\pi i}{n!} f^{(n)}(a)$ . That $1/n!$ is the most commonly dropped factor in this whole topic.

For a general integrand, classify each isolated singularity from its Laurent series. A removable singularity has no negative powers, a pole of order $m$ has lowest term $(z-a)^{-m}$ , and an essential singularity has infinitely many negative powers. At a simple pole, $\operatorname{Res}_{z=a} f = \lim_{z \to a}(z-a)f(z)$ , and when $f = p/q$ with $q$ having a simple zero at $a$ and $p(a) \neq 0$ , this reduces to $p(a)/q'(a)$ . At a pole of order $m$ , differentiate $m-1$ times: $\operatorname{Res}_{z=a} f = \dfrac{1}{(m-1)!}\lim_{z\to a}\dfrac{d^{m-1}}{dz^{m-1}}\left[(z-a)^m f(z)\right]$ . Sum the residues of the enclosed singularities only, then multiply by $2\pi i$ .

After this page, you should be able to

Test a candidate function for analyticity with the Cauchy-Riemann equations and say why $\bar{z}$ fails.
Decide when Cauchy's theorem alone forces a contour integral to vanish.
Classify an isolated singularity as removable, a pole of stated order, or essential.
Compute residues at simple poles and at poles of higher order with the correct formula.
Evaluate a closed contour integral by summing only the residues enclosed by the contour.

Formulas and assumptions

Cauchy-Riemann equations

u_x = v_y, \qquad u_y = -v_x, \qquad f'(z) = u_x + i v_x

Variables

u: the real part of f as a function of x and y
v: the imaginary part of f

Assumptions

Necessary for analyticity; sufficient when u and v have continuous partial derivatives near the point.

Cauchy's integral formula for derivatives

\oint_{C} \frac{f(z)}{(z-a)^{n+1}}\,dz = \frac{2\pi i}{n!}\, f^{(n)}(a)

Variables

f: analytic on and inside the simple closed contour C
a: a point enclosed by C
n: a nonnegative integer

Assumptions

C is a simple closed contour traversed once counterclockwise and a lies inside it.
n = 0 recovers the ordinary Cauchy integral formula.

Residue at a pole of order m

\operatorname*{Res}_{z=a} f = \frac{1}{(m-1)!}\lim_{z \to a} \frac{d^{\,m-1}}{dz^{\,m-1}}\left[(z-a)^{m} f(z)\right]

Variables

a: the location of the pole
m: the order of the pole

Assumptions

The singularity is a pole of finite order m, not essential.
For m = 1 this is the limit of (z - a) f(z).

Residue theorem

\oint_{C} f(z)\,dz = 2\pi i \sum_{k} \operatorname*{Res}_{z=a_k} f

Variables

C: a simple closed contour traversed once counterclockwise
a_k: the isolated singularities of f enclosed by C

Assumptions

f is analytic on C and inside it except at the isolated singularities a_k.
Singularities outside C contribute nothing.

Worked example

A contour integral with two simple poles

Evaluate $\displaystyle \oint_{|z| = 2} \frac{z^{2}+1}{z(z-1)}\,dz$ , where the circle is traversed once counterclockwise.

1The denominator vanishes at $z = 0$ and $z = 1$ . Each is a simple zero, and the numerator is nonzero there ( $1$ and $2$ respectively), so both are simple poles.
2Both lie inside the contour: $|0| = 0 < 2$ and $|1| = 1 < 2$ , so both residues count.
3Residue at $z = 0$ : $\displaystyle \lim_{z\to 0} z \cdot \frac{z^{2}+1}{z(z-1)} = \left.\frac{z^{2}+1}{z-1}\right|_{z=0} = \frac{1}{-1} = -1$ .
4Residue at $z = 1$ : $\displaystyle \lim_{z\to 1} (z-1)\cdot \frac{z^{2}+1}{z(z-1)} = \left.\frac{z^{2}+1}{z}\right|_{z=1} = \frac{2}{1} = 2$ .
5Apply the residue theorem: $\oint = 2\pi i\left(-1 + 2\right) = 2\pi i$ .

$\displaystyle \oint_{|z|=2} \frac{z^{2}+1}{z(z-1)}\,dz = 2\pi i$ .

Common traps

Including residues at singularities that lie outside the contour, or missing one that lies inside — always compare $|a_k|$ with the contour radius first.
Dropping the $1/n!$ in Cauchy's integral formula for derivatives, which scales the answer by a factor of $n!$ .
Using the simple-pole limit $\lim (z-a)f(z)$ at a higher-order pole, where that limit is infinite and the order- $m$ formula is required.
Concluding a function is analytic because the Cauchy-Riemann equations hold at one point; analyticity requires them on a neighbourhood, with the partials continuous.
Assuming $\bar{z}$ , $|z|$ , and $\operatorname{Re} z$ are analytic because they look smooth — none of them satisfies the Cauchy-Riemann equations.

A Contour Integral at a Third-Order Pole GRE Mathematics overview Series Convergence Tests Limits and Continuity Sign in for current practice

Question depth and domain coverage vary by exam. Practice answers are checked after submission.

Sources

GRE Subject Test Content and Structure — ETS. Accessed 2026-07-06. Use as a cited source for exam facts; do not imply affiliation or reproduce protected test material.
Complex Variables with Applications (Orloff), Section 2.6: Cauchy-Riemann Equations — LibreTexts Mathematics. Accessed 2026-08-15. Orloff's Complex Variables with Applications on LibreTexts is CC BY-NC-SA 4.0; attribute the author and platform and avoid verbatim reuse beyond short cited references.
Complex Variables with Applications (Orloff), Section 4.6: Cauchy's Theorem — LibreTexts Mathematics. Accessed 2026-08-15. Orloff's Complex Variables with Applications on LibreTexts is CC BY-NC-SA 4.0; attribute the author and platform and avoid verbatim reuse beyond short cited references.
Complex Variables with Applications (Orloff), Section 5.1: Cauchy's Integral for Functions — LibreTexts Mathematics. Accessed 2026-08-15. Orloff's Complex Variables with Applications on LibreTexts is CC BY-NC-SA 4.0; attribute the author and platform and avoid verbatim reuse beyond short cited references.
Complex Variables with Applications (Orloff), Section 5.2: Cauchy's Integral Formula for Derivatives — LibreTexts Mathematics. Accessed 2026-08-15. Orloff's Complex Variables with Applications on LibreTexts is CC BY-NC-SA 4.0; attribute the author and platform and avoid verbatim reuse beyond short cited references.
Complex Variables with Applications (Orloff), Section 8.7: Laurent Series — LibreTexts Mathematics. Accessed 2026-08-15. Orloff's Complex Variables with Applications on LibreTexts is CC BY-NC-SA 4.0; attribute the author and platform and avoid verbatim reuse beyond short cited references.
Complex Variables with Applications (Orloff), Section 9.4: Residues — LibreTexts Mathematics. Accessed 2026-08-15. Orloff's Complex Variables with Applications on LibreTexts is CC BY-NC-SA 4.0; attribute the author and platform and avoid verbatim reuse beyond short cited references.
Complex Variables with Applications (Orloff), Section 9.5: Cauchy Residue Theorem — LibreTexts Mathematics. Accessed 2026-08-15. Orloff's Complex Variables with Applications on LibreTexts is CC BY-NC-SA 4.0; attribute the author and platform and avoid verbatim reuse beyond short cited references.

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Analytic Functions and Residues

Concise answer

Definitions

Intuition

Concept walkthrough

After this page, you should be able to

Formulas and assumptions

Cauchy-Riemann equations

Cauchy's integral formula for derivatives

Residue at a pole of order m

Residue theorem

Worked example

A contour integral with two simple poles

Common traps

Related pages and practice

Sources

Sources and corrections