Complex Variables Questions and Answers

Answer: D

The real line is closed by a semicircle in the upper half plane; if the integrand decays fast enough, that arc contributes nothing in the limit and the real integral equals \( 2\pi i \) times the residues in the upper half plane. A small circle round the origin is the tool for a principal value, not for the whole line

Answer: C

A removable singularity is one the expansion does not notice, having no negative powers, so a value can simply be assigned. A pole of order \( m \) is bounded below at \( (z-z_{0})^{-m} \). An essential singularity has no such floor, which is why the function has no limit there in any direction

Answer: B

A translation moves the centre without changing the radius. A real scaling multiplies the radius. Inversion sends \( |z| = 1 \) to \( |w| = 1 \), so the unit circle is carried onto itself, though each point moves to its own conjugate reflection round it

Answer: A

The complex integrand and contour come first. The arc estimate must be settled next, because without it the closed-contour value says nothing about the real integral. Only then are the residues worth computing, and the factor \( 2\pi i \) applied at the end

Answer: D

The first and third hold, the third being one of the striking differences from real analysis. The second omits a hypothesis: the equations must be accompanied by continuity of the partial derivatives, without which functions exist that satisfy them at a point yet are not differentiable there

Answer: C

The first and third hold. The second reverses the implication: the integral of \( 1/(z^{2}-1) \) round \( |z| = 2 \) vanishes although two poles lie inside, the residues merely cancelling. A vanishing integral is therefore evidence of nothing on its own

Answer: B

The exponential satisfies the Cauchy-Riemann equations at every point and is its own derivative, so it is entire. The conjugate fails them everywhere, the reciprocal fails at the origin, and the modulus is real-valued and non-constant, which no analytic function can be

Answer: A

The maximum modulus principle forbids an interior maximum unless the function is constant: the largest modulus is taken on the boundary. The Taylor expansion, the vanishing contour integral on a simply connected region, and the harmonicity of the real part are all genuine consequences of analyticity

Answer: D

Harmonicity of each part is necessary but not sufficient: the two must be conjugate, that is linked by the Cauchy-Riemann equations. Taking \( u = x \) with \( v = y \) gives two harmonic functions, yet \( x + iy \) written that way is \( \bar{z} \)'s cousin and fails the equations, so the pairing matters as much as the parts

Answer: C

For a pole of order two the recipe is

\[ \mathrm{Res}_{z=a} f \;=\; \lim_{z \to a} \frac{d}{dz}\left[(z-a)^{2}f(z)\right], \]

so the squared factor removes the pole and the differentiation extracts the coefficient \( a_{-1} \) rather than \( a_{-2} \). A double pole does have a residue, and it may well be complex