A function has a Laurent expansion around z₀ where the principal part contains exactly three terms (down to (z − z₀)^{−3}). What type of singularity does z₀ represent?
AAn essential singularity, because negative powers are present
BA removable singularity, because the series converges near z₀
CA pole of order 3, because the principal part terminates at (z − z₀)^{−3}
DA branch point, because the index goes negative
A singularity is a pole of order m when the principal part has finitely many terms, terminating at (z − z₀)^{−m}. Three terms ending at the −3 power means a pole of order 3. An essential singularity requires infinitely many terms in the principal part. Option A is the classic misconception — the mere presence of negative powers does not imply an essential singularity; it only does when those terms continue without end.
Question 2 Multiple Choice
Why is the coefficient a₋₁ in a Laurent series singled out as 'the residue'?
AIt is always the largest coefficient and dominates near the singularity
BIt is the only Laurent coefficient that survives integration around a closed loop encircling z₀
CIt equals the limit of f(z) as z → z₀, giving the singularity's strength
DIt determines whether the singularity is removable or not
Integrating (z − z₀)^n around a small loop gives zero for every integer n except n = −1, where it gives 2πi. Therefore (1/2πi)∮f(z)dz = a₋₁ — only the (z − z₀)^{−1} term contributes. All other terms integrate to zero around a closed path. This is why the residue is the key number for computing contour integrals, not because it is necessarily the largest or limit-related coefficient.
Question 3 True / False
The principal part of a Laurent series converges on the same disk-shaped region as the regular (non-negative power) part.
TTrue
FFalse
Answer: False
The two parts converge on different regions. The regular part Σ_{n≥0} aₙ(z−z₀)^n converges inside a disk |z − z₀| < R. The principal part Σ_{n<0} aₙ(z−z₀)^n is a power series in 1/(z−z₀) that converges outside a circle |z − z₀| > r. The Laurent series as a whole converges on the annulus r < |z − z₀| < R where both parts converge simultaneously — not on a disk.
Question 4 True / False
Every singularity of a complex function can be classified by examining how many terms appear in the principal part of its Laurent series.
TTrue
FFalse
Answer: True
The classification is complete and based solely on the principal part: zero terms (principal part is empty) means a removable singularity; finitely many terms ending at (z−z₀)^{−m} means a pole of order m; infinitely many terms means an essential singularity. The regular part plays no role in the classification — only the negative-power terms matter.
Question 5 Short Answer
Explain why a₋₁ — the residue — has special significance for contour integration, while no other Laurent coefficient does.
Think about your answer, then reveal below.
Model answer: When you integrate (z − z₀)^n around a closed loop encircling z₀, the result is 2πi if n = −1 and zero for all other integers n. So integrating term-by-term through the Laurent series, only the (z − z₀)^{−1} term survives. All other terms contribute nothing. The residue a₋₁ is therefore the sole Laurent coefficient that encodes what a contour integral 'sees' at a singularity, which is why residues are the central tool for evaluating complex integrals.
This is a consequence of the antiderivative: (z − z₀)^n for n ≠ −1 has an antiderivative (z − z₀)^{n+1}/(n+1) that returns to its starting value after a full loop, giving zero net integral. The n = −1 case is the exception because log(z − z₀) is multivalued — it does not return to its original value after encircling z₀, picking up the factor 2πi instead.