Complex Variables Questions and Answers

Answer: B

The formula reads

\[ f(a) \;=\; \frac{1}{2\pi i}\oint_{C} \frac{f(z)}{z - a}\,dz, \]

so the values on the contour determine the value at every interior point. The factor \( 2\pi i \) belongs in the denominator, and the \( i \) with it: without the \( i \) the formula would not even be dimensionally consistent as a complex identity

Answer: A

Put \( z = e^{i\theta} \), so \( dz = ie^{i\theta}d\theta \):

\[ \oint_{|z|=1}\frac{dz}{z} \;=\; \int_{0}^{2\pi} \frac{ie^{i\theta}}{e^{i\theta}}\,d\theta \;=\; i\int_{0}^{2\pi} d\theta \;=\; 2\pi i. \]

It is non-zero because the origin, where the integrand fails to be analytic, lies inside the contour

Answer: D

A Laurent series runs over all integer powers, \( \sum_{n=-\infty}^{\infty} a_{n}(z-z_{0})^{n} \), and converges in an annulus rather than a disc. The negative powers are what let it represent a function round an isolated singularity, and the coefficient \( a_{-1} \) is the residue there

Answer: C

A removable singularity has no negative powers, a pole of order \( m \) has finitely many down to \( (z - z_{0})^{-m} \), and an essential singularity has infinitely many. The standard example is \( e^{1/z} \) at the origin, near which the function takes every value infinitely often, so no limit exists in any sense

Answer: B

Expanding the numerator and dividing,

\[ \frac{\sin z}{z} \;=\; \frac{z - z^{3}/6 + \cdots}{z} \;=\; 1 - \frac{z^{2}}{6} + \cdots, \]

a series with no negative powers at all. The limit is one, so assigning that value at the origin makes the function analytic there. The zero of the numerator cancels the zero of the denominator

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

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

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

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

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 \( \pm\pi i \)