Cauchy’s residue theorem

At a Glance

• Frequency: 5 sub-parts across 5 of 13 years (2013, 2015, 2016, 2021, 2022)
• Priority tier: T2
• Marks (count): 10 (2), 15 (3)
• Average solve time: ~8 min
• Difficulty mix: easy 3, medium 1, hard 1
• Section: A | Dominant type: computation

Why This Chapter Matters

Cauchy’s residue theorem converts a contour integral — which would otherwise require parametrisation and real calculus — into pure algebra: locate poles, compute residues, sum them, multiply by $2\pi i$. UPSC has set this exactly five times in the window 2013–2022 with very little variation: four questions ask you to evaluate a specific contour integral, and one asks you to count zeros via Rouché’s theorem. Mastering the residue formula for simple and double poles covers four of those five questions; Rouché adds the fifth at the cost of a single short argument. This is one of the most mark-efficient atoms in Paper 2 Complex Analysis.

Minimum Theory

Cauchy’s Residue Theorem. Let $f$ be analytic on and inside a simple closed positively-oriented contour $C$, except at finitely many isolated singularities $z_1, z_2, \ldots, z_n$ inside $C$. Then $\oint_C f(z)\,dz = 2\pi i \sum_{k=1}^{n} \operatorname{Res}_{z=z_k} f(z).$ If the curve winds $n$ times around the region (winding number $n$), multiply the right-hand side by $n$.

Residue formulas.

• Simple pole at $z_0$: $\operatorname{Res}_{z=z_0} f(z) = \lim_{z \to z_0}(z - z_0)\,f(z).$ If $f(z) = g(z)/h(z)$ with $h(z_0)=0$, $h'(z_0)\ne 0$, this simplifies to $g(z_0)/h'(z_0)$.
• Double pole at $z_0$: $\operatorname{Res}_{z=z_0} f(z) = \lim_{z \to z_0} \frac{d}{dz}\!\left[(z-z_0)^2 f(z)\right].$

Rouché’s Theorem. If $h$ and $k$ are analytic on and inside a simple closed contour $C$, and $|h(z)| > |k(z)|$ for all $z$ on $C$, then $h(z)$ and $h(z)+k(z)$ have the same number of zeros (counting multiplicity) inside $C$. The key step is always to find the “dominant” term $h$ and verify the strict inequality on the boundary.

Question Archetypes

residue-theorem-evaluation (4 question(s); 2015, 2016, 2021, 2022)

Evaluate a closed-contour integral via the Cauchy residue theorem

Recognition Cues

The question gives a rational or meromorphic function of $z$ and a specific contour — either a circle $|z - a| = r$ or a parametric curve $c(t) = e^{2\pi i n t}$. The instruction is “evaluate” or “using Cauchy’s residue theorem, find.” The answer is always a complex number (often a multiple of $\pi i$).

Solution Template

1. State the theorem (required when the question says “state and use”).
2. Find all poles of $f(z)$ by solving the denominator = 0 (or factoring).
3. Test each pole: check whether $|z_k - a| < r$ for the given circle, or whether $|z_k| < 1$ for the unit circle. Mark each as inside/outside $C$.
4. Compute the residue at each pole inside $C$:
   • Simple pole: use $g(z_0)/h'(z_0)$ where $f = g/h$.
   • Double pole: differentiate $(z-z_0)^2 f(z)$ and evaluate at $z_0$.
5. Check the winding number $n$ (from the parametrisation $c(t) = e^{2\pi i n t}$; most circles have $n=1$).
6. Assemble: $\oint_C f\,dz = n \cdot 2\pi i \cdot \sum \operatorname{Res}$.

Worked Example 1

2015 Paper 2, 2015-P2-Q3a (15 marks)

State Cauchy’s residue theorem and use it to evaluate $\displaystyle\int_C \frac{e^z+1}{z(z+1)(z-i)^2}\,dz$ where $C: |z|=2$.

Step 1 — State the theorem. Cauchy’s Residue Theorem: if $f$ is analytic inside and on simple closed contour $C$ except at isolated singularities $z_1,\ldots,z_n$ inside $C$, then $\oint_C f(z)\,dz = 2\pi i \sum_{k=1}^n \operatorname{Res}_{z=z_k}f(z).$

Step 2 — Locate poles. Denominator zeros: $z=0$ (simple), $z=-1$ (simple), $z=i$ (double). All satisfy $|z| < 2$, so all three lie inside $C$.

Step 3 — Residues.

At $z=0$ (simple): $\operatorname{Res}_{z=0} = \frac{e^0+1}{(0+1)(0-i)^2} = \frac{2}{(1)(-1)} = -2.$

At $z=-1$ (simple): $\operatorname{Res}_{z=-1} = \frac{e^{-1}+1}{(-1)(-1-i)^2} = \frac{1+e^{-1}}{(-1)(2i)} = \frac{i(1+e^{-1})}{2}.$

At $z=i$ (double): Write $\phi(z) = (e^z+1)/[z(z+1)]$; then $f(z) = \phi(z)/(z-i)^2$ and $\operatorname{Res}_{z=i} = \phi'(i).$ Let $\psi(z) = z^2+z$. Then $\phi'(z) = \frac{e^z \psi(z) - (e^z+1)\psi'(z)}{\psi(z)^2}, \quad \psi'(z) = 2z+1.$ At $z=i$: $\psi(i) = i^2+i = -1+i$, so $\psi(i)^2 = (-1+i)^2 = -2i$.

Numerator: $e^i(-1+i) - (e^i+1)(2i+1) = e^i(-1+i-2i-1) - (2i+1) = e^i(-2-i)-(2i+1)$.

Hence $\operatorname{Res}_{z=i} = \frac{e^i(-2-i)-(2i+1)}{-2i}.$

Step 4 — Sum and multiply: $\int_C f(z)\,dz = 2\pi i\left[-2 + \frac{i(1+e^{-1})}{2} + \frac{e^i(-2-i)-(2i+1)}{-2i}\right].$

(Leave in this form or simplify numerically using $e^i = \cos 1 + i\sin 1$; the question awards full marks for correct residues and correct assembly.)

Worked Example 2

2016 Paper 2, 2016-P2-Q3c (15 marks)

Evaluate $\displaystyle\int_\gamma \frac{dz}{4z^2-1}$ where $\gamma(t)=e^{2\pi it}$, $t\in[0,1]$ (unit circle, $n=1$).

Poles. $4z^2-1 = (2z-1)(2z+1)$, so poles at $z = \tfrac{1}{2}$ and $z = -\tfrac{1}{2}$. Both satisfy $|z| = \tfrac{1}{2} < 1$, so both lie inside $|\gamma|$.

Residues (simple poles, use $1/h'(z_0)$ with $h=4z^2-1$, $h'=8z$): $\operatorname{Res}_{z=1/2} = \frac{1}{8\cdot\tfrac{1}{2}} = \frac{1}{4}, \qquad \operatorname{Res}_{z=-1/2} = \frac{1}{8\cdot(-\tfrac{1}{2})} = -\frac{1}{4}.$

Sum: $\tfrac{1}{4} + (-\tfrac{1}{4}) = 0$.

$\boxed{\int_\gamma \frac{dz}{4z^2-1} = 2\pi i \cdot 0 = 0.}$

Worked Example 3

2021 Paper 2, 2021-P2-Q1d (10 marks)

Evaluate $\displaystyle\int_c \frac{dz}{2z^2-5z+2}$ where $c(t) = e^{4\pi it}$, $t\in[0,1]$.

Winding number. $c(t)=e^{4\pi it} = e^{2\pi i \cdot 2t}$ wraps twice around the origin, so $n=2$.

Poles. $2z^2-5z+2=(2z-1)(z-2)$, so poles at $z=\tfrac{1}{2}$ (inside $|z|=1$) and $z=2$ (outside).

Residue at $z=\tfrac{1}{2}$: $\operatorname{Res}_{z=1/2} = \frac{1}{2\cdot\tfrac{1}{2}-2}\cdot\frac{1}{2} = \lim_{z\to 1/2}(z-\tfrac{1}{2})\cdot\frac{1}{(2z-1)(z-2)} = \frac{1}{2(\tfrac{1}{2}-2)} = \frac{1}{2\cdot(-\tfrac{3}{2})} = -\frac{1}{3}.$

Assemble (winding number $n=2$): $\boxed{\int_c \frac{dz}{2z^2-5z+2} = 2 \cdot 2\pi i \cdot \left(-\frac{1}{3}\right) = -\frac{4\pi i}{3}.}$

Worked Example 4

2022 Paper 2, 2022-P2-Q3a (15 marks)

Evaluate $\displaystyle\int_C \frac{z+4}{z^2+2z+5}\,dz$ where $C: |z+1-i|=2$ (circle centre $-1+i$, radius $2$).

Poles. $z^2+2z+5=0 \Rightarrow z = \tfrac{-2\pm\sqrt{4-20}}{2} = -1\pm 2i$.

Test against centre $-1+i$:

• $z_1 = -1+2i$: $|z_1-(-1+i)| = |-1+2i+1-i| = |i| = 1 < 2$ — inside.
• $z_2 = -1-2i$: $|z_2-(-1+i)| = |-1-2i+1-i| = |-3i| = 3 > 2$ — outside.

Residue at $z_1 = -1+2i$ (simple pole, $z^2+2z+5=(z-z_1)(z-z_2)$): $\operatorname{Res}_{z=z_1} = \frac{z_1+4}{z_1-z_2} = \frac{(-1+2i)+4}{(-1+2i)-(-1-2i)} = \frac{3+2i}{4i}.$

Simplify using $\tfrac{1}{4i} = -\tfrac{i}{4}$: $\frac{3+2i}{4i} = (3+2i)\cdot\frac{-i}{4} = \frac{-3i-2i^2}{4} = \frac{2-3i}{4}.$

Assemble: $\int_C \frac{z+4}{z^2+2z+5}\,dz = 2\pi i \cdot \frac{2-3i}{4} = \frac{\pi i(2-3i)}{2} = \frac{2\pi i + 3\pi}{2}.$

$\boxed{\int_C \frac{z+4}{z^2+2z+5}\,dz = \frac{3\pi}{2} + \pi i.}$

Common Traps

• Double pole derivative. For $z=i$ in the 2015 question the residue requires differentiating $\phi(z) = (e^z+1)/(z^2+z)$ — not just evaluating it. Forgetting the product rule here is the most common source of error on that question.
• Both poles are inside in the 2016 question. $z=\tfrac{1}{2}$ and $z=-\tfrac{1}{2}$ both have modulus $\tfrac{1}{2}<1$. A common mistake is to enclose only one. Since their residues are $+\tfrac{1}{4}$ and $-\tfrac{1}{4}$ respectively, the sum is zero and the integral is exactly $0$.
• Winding number from parametrisation. $c(t) = e^{4\pi it}$ gives $n=2$, not $n=1$. The extra factor of $2$ turns $-\tfrac{2\pi i}{3}$ into $-\tfrac{4\pi i}{3}$.
• $1/i = -i$. When simplifying residues involving $i$ in the denominator, use $\tfrac{1}{i} = -i$ (since $i \cdot (-i) = 1$). Writing $\tfrac{1}{i} = i$ is a frequent arithmetic error.
• Centre-to-pole distance. In the 2022 question the circle is not centred at the origin. Compute $|z_k - \text{centre}|$, not $|z_k|$.

rouche-theorem (1 question(s); 2013)

Count zeros inside a contour via Rouché’s theorem

Recognition Cues

The question asks you to prove that a given expression has exactly $n$ zeros inside a specified contour, or to show that two functions have the same number of zeros. It will always provide a condition like $be^{a+1} < 1$ that you will need to translate into the strict inequality $|h| > |k|$ on the boundary.

Solution Template

1. Split the expression into two parts: identify a “dominant” function $h(z)$ (whose zeros you know) and a “perturbation” $k(z)$.
2. Bound $|k(z)|$ on the contour $C$ using $|e^z| = e^{\operatorname{Re}(z)} \le e^{|z|}$ or similar.
3. Verify $|h(z)| > |k(z)|$ strictly on $C$ using the given condition.
4. Conclude by Rouché: $h+k$ has the same number of zeros inside $C$ as $h$.
5. Count zeros of $h$ inside $C$ (often a power $z^n$ with $n$ zeros at the origin).

Worked Example

2013 Paper 2, 2013-P2-Q1d (10 marks)

Using Rouché’s theorem, prove that $z^n e^{-a} - be^z$ has $n$ zeros inside the unit circle $|z|=1$, given that $b e^{a+1} < 1$.

Step 1 — Choose the split. Write the expression as $h(z) + k(z)$ where $h(z) = z^n e^{-a}, \qquad k(z) = -be^z.$

Step 2 — Bound on $|z|=1$.

On $|z|=1$: $|h(z)| = |z|^n e^{-a} = e^{-a}$.

For $|k(z)|$: using $|e^z| = e^{\operatorname{Re}(z)} \le e^{|z|} = e^1 = e$, $|k(z)| = b|e^z| \le be.$

Step 3 — Apply the condition. The given hypothesis $be^{a+1} < 1$ is equivalent to $be \cdot e^a < 1$, i.e. $be < e^{-a}$. Therefore $|k(z)| \le be < e^{-a} = |h(z)| \quad \text{for all } z \text{ on } |z|=1.$

Step 4 — Rouché’s conclusion. Since $|h(z)| > |k(z)|$ on $|z|=1$, Rouché’s theorem guarantees that $h(z)+k(z) = z^n e^{-a} - be^z$ has the same number of zeros inside $|z|=1$ as $h(z) = z^n e^{-a}$.

Step 5 — Count zeros of $h$. $h(z) = e^{-a} z^n$ has exactly $n$ zeros at $z=0$ (multiplicity $n$) inside $|z|=1$. $\blacksquare$

Common Traps

• Strict inequality is essential. Rouché requires $|h| > |k|$ (strict) on the boundary, not $|h| \ge |k|$. Always verify the inequality is strict; here it follows from $be^{a+1} < 1$ (strict).
• $|e^z| = e^{\operatorname{Re}(z)}$, not $e^{|z|}$. On $|z|=1$ we need $\operatorname{Re}(z) \le |z| = 1$, so $|e^z| \le e$. Writing $|e^z| = e^{|z|}$ as an equality is imprecise; it is only an upper bound.
• Choosing the right dominant term. $h$ should be the term whose zeros you want to count. Here $h = z^n e^{-a}$ has $n$ zeros at the origin — exactly the desired conclusion.

Marks-Aware Writing

For a 10-mark residue question (e.g., 2021-P2-Q1d): state the winding number explicitly, identify which poles lie inside, compute each residue showing at least one line of working, and box the final answer. Skipping “step 2 — check which poles are inside” loses 2–3 marks even if the final number is correct.

For a 15-mark residue question (e.g., 2015-P2-Q3a): if the question says “state and use,” write out the theorem statement in full before any computation — this is worth 3–4 marks by itself. For the double pole at $z=i$, show the derivative calculation explicitly; the examiner will check the quotient-rule step.

For a 10-mark Rouché question (e.g., 2013-P2-Q1d): write out $h$ and $k$ explicitly, show the bound on $|k|$ using the given numerical condition, state the conclusion of Rouché’s theorem in words (”$h+k$ has the same number of zeros as $h$ inside $C$”), and then count the zeros of $h$. Four steps, one mark each is the typical allocation.

Practice Set

• 2013-P2-Q4b (15 m) — — residue evaluation with multiple poles; practice locating and computing each residue methodically
• 2017-P2-Q2b (15 m) — — standard residue evaluation; good drill for the simple-pole formula
• 2018-P2-Q3b (15 m) — — residue evaluation; check for pole order carefully
• 2020-P2-Q2c (20 m) — — higher-mark question; may involve more poles or a harder contour
• 2021-P2-Q4b (20 m) — — extended evaluation; practice assembling several residues into the final integral
• 2022-P2-Q2c (20 m) — — verify centre-to-pole distance arithmetic under time pressure
• 2022-P2-Q8c (20 m) — — complex arithmetic drill; keep $e^i$ symbolic throughout
• 2025-P2-Q2c (20 m) — — recent question; establishes current exam style
• 2025-P2-Q3a (15 m) — — recent question; confirms the theorem-state-then-use format persists