Determine the order of the pole of \( \dfrac{1}{(z-1)^{3}} \) at \( z = 1 \).
Answer: A
The order of a pole is the power of the vanishing factor in the denominator, here three, so the Laurent expansion about \( z = 1 \) runs down to \( (z-1)^{-3} \) and no further. A first power would give a simple pole, and no finite power gives an essential singularity
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For a simple pole at \( z = a \), the residue of \( f \) is
Answer: D
Multiplying by the vanishing factor cancels the pole and leaves the coefficient \( a_{-1} \):
\[ \mathrm{Res}_{z=a} f \;=\; \lim_{z \to a} (z - a)f(z). \]
The square would be needed for a double pole, together with one differentiation, and the plain limit of \( f \) does not exist at a pole at all
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Find the residue of \( \dfrac{1}{z^{2} - 1} \) at \( z = 1 \).
Answer: C
Factorise and take the limit:
\[ \mathrm{Res}_{z=1} \;=\; \lim_{z \to 1} (z-1)\cdot\frac{1}{(z-1)(z+1)} \;=\; \frac{1}{1 + 1} \;=\; \frac{1}{2}. \]
The residue at the other pole \( z = -1 \) works out at \( -1/2 \), and the two sum to zero, which is why a contour enclosing both gives a vanishing integral
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The residue theorem gives \( \displaystyle\oint_{C} f\,dz \) as
Answer: B
The theorem reads
\[ \oint_{C} f(z)\,dz \;=\; 2\pi i \sum_{k} \mathrm{Res}_{z=z_{k}} f, \]
summed over the singularities inside \( C \). Cauchy's theorem is the special case of no enclosed singularity, and the factor of \( i \) traces back to \( \oint dz/z = 2\pi i \) round a single simple pole
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Taken anticlockwise round the circle of radius two, the integral \( \oint dz/(z^{2}-1) \) comes to
Answer: A
Both poles \( z = \pm 1 \) lie inside \( |z| = 2 \), with residues \( 1/2 \) and \( -1/2 \), so
\[ \oint \frac{dz}{z^{2}-1} \;=\; 2\pi i\left(\tfrac{1}{2} - \tfrac{1}{2}\right) \;=\; 0. \]
The vanishing is a cancellation between the two residues, not a consequence of analyticity: a contour enclosing only one of them would give \( \pm\pi i \)
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Real integrals over \( (-\infty, \infty) \) are evaluated by residues using a contour that is
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
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Match the singularity with its Laurent expansion. 1. Removable 2. Pole of order m 3. Essential P. Negative powers down to the m-th Q. No negative powers R. Infinitely many negative powers
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
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Match the map with what it does to the unit circle. 1. \( w = z + 1 \) 2. \( w = 2z \) 3. \( w = 1/z \) P. Doubles its radius Q. Shifts its centre R. Leaves it fixed as a set
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
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Put these steps of evaluating a real integral by residues in order. 1. Sum the residues of the poles in the upper half plane 2. Show the semicircular arc contributes nothing in the limit 3. Multiply by \( 2\pi i \) and take the real part if required 4. Replace the real integrand by a function of a complex variable and choose the contour
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
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Consider the following statements about analytic functions. 1. The real and imaginary parts are both harmonic 2. The Cauchy-Riemann equations alone guarantee analyticity 3. An analytic function is infinitely differentiable
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
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