Complex Variables Questions and Answers

Answer: C

The derivative is \( 2z \), which vanishes only at the origin. There the map doubles angles rather than preserving them: a right angle between the axes becomes a straight angle, since \( \arg w = 2\arg z \). Such a point is called critical, and conformality is lost exactly where the derivative is zero

Answer: B

A bilinear map is a composition of translations, scalings, rotations and the inversion \( z \mapsto 1/z \), and each of those carries the family of circles-and-lines to itself. Treating a line as a circle through the point at infinity makes the statement exact, and the condition \( ad - bc \neq 0 \) is what stops the map collapsing to a constant

Answer: A

Setting \( w = z \) gives

\[ cz^{2} + (d - a)z - b \;=\; 0, \]

a quadratic, so there are at most two roots. A repeated root gives one fixed point, and the identity is the degenerate case fixing everything. Being determined by three point-images is a separate property, and is where the count of three belongs

Answer: D

Taking moduli, \( |w| = 1/|z| \), so

\[ |z| = 2 \quad \Longrightarrow \quad |w| \;=\; \tfrac{1}{2}. \]

The inversion turns the circle inside out about the unit circle, which is left fixed. A circle maps to a straight line only when it passes through the origin, where the map sends points to infinity

Answer: C

Analyticity throughout the region enclosed, with no singularity inside, forces the integral round any closed contour to vanish. Continuity or boundedness is far too weak: \( 1/z \) is bounded and continuous on \( |z| = 1 \) yet integrates to \( 2\pi i \), because it fails to be analytic at the enclosed origin

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