Contour integral of a function with two enclosed poles
BlinkNBuild practice problem · GATE standard, authored and verified in-house
Evaluate $$\oint_C \frac{dz}{z^2 + 4},$$ where $C$ is the circle $|z| = 3$ traversed counter-clockwise.
The value of the integral is ________.
Show the step-by-step derivation
Answer
0 exactly zero exactly zero
Step-by-step derivation
- Factor the denominator to locate the singularities: $$z^2 + 4 = (z - 2i)(z + 2i),$$ so there are simple poles at $z = 2i$ and $z = -2i$.
- Check which lie inside the contour. Both have modulus $|{\pm}2i| = 2 < 3$, so both poles are enclosed by $|z| = 3$.
- Compute the residue at $z = 2i$. For a simple pole of $f = p/q$, $\text{Res} = p/q'$ with $q' = 2z$: $$\text{Res}_{z=2i} = \frac{1}{2(2i)} = \frac{1}{4i}.$$
- Compute the residue at $z = -2i$: $$\text{Res}_{z=-2i} = \frac{1}{2(-2i)} = -\frac{1}{4i}.$$
- Sum the residues: $$\frac{1}{4i} - \frac{1}{4i} = 0.$$ The two residues are exact negatives of one another.
- Apply the residue theorem: $$\oint_C \frac{dz}{z^2+4} = 2\pi i \sum \text{Res} = 2\pi i \times 0 = \mathbf{0}.$$
- Why the cancellation is not a coincidence. The poles are complex conjugates with conjugate residues, and enclosing both makes the imaginary parts cancel. Shrink the contour to $|z| = 1$ and neither pole is inside, so the integral is $0$ again - but for a completely different reason. Enclose only one, say with a contour around $z = 2i$ alone, and the answer becomes $2\pi i/(4i) = \pi/2 \ne 0$.
The trap this question is built around
Assuming a non-zero answer because poles are enclosed. The residue theorem sums residues, and a sum can vanish. Always compute both and add - checking only whether poles are inside is half the work.
The idea behind this question
The residue theorem turns a contour integral into a sum: $2\pi j$ times the sum of the residues of the poles inside the contour. Residues can be positive or negative, complex or real, so poles inside do not guarantee a non-zero result - sometimes they cancel exactly.
Try a variation
Evaluate $\oint_C \dfrac{z\,dz}{z^2 + 4}$ on the same circle $|z| = 3$.
Show the answer
Answer: $2\pi j$
Each pole $\pm 2j$ has residue $\tfrac12$, so the sum is $1$ and the integral $2\pi j$.
Other mistakes to avoid
- Including a pole that lies outside the contour.
- Forgetting the factor $2\pi j$, or its sign for a clockwise contour.