Complex Variables Questions and Answers

Answer: A

Complex differentiability at an isolated point is too weak to be useful; analyticity demands it on a whole open disc about the point. That extra requirement is what buys the powerful results -- infinite differentiability, a convergent Taylor series, and Cauchy's theorem -- none of which follows from differentiability at one point alone

Answer: D

Requiring the derivative to be the same whether the limit is taken along the real or the imaginary direction gives

\[ \frac{\partial u}{\partial x} \;=\; \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} \;=\; -\frac{\partial v}{\partial x}. \]

The minus sign belongs to the second pair only; putting it on the first, or omitting it altogether, describes no analytic function

Answer: C

Here \( u = x \) and \( v = -y \), so \( u_{x} = 1 \) while \( v_{y} = -1 \):

\[ u_{x} \;=\; 1 \;\neq\; -1 \;=\; v_{y}. \]

The first Cauchy-Riemann equation fails at every point, so the conjugate is nowhere analytic despite being continuous everywhere. Continuity is no guarantee of complex differentiability

Answer: B

The equations are necessary, and they become sufficient once the four partial derivatives are also continuous at the point. Without that extra hypothesis there are functions satisfying them at the origin yet not differentiable there, so the equations alone cannot be treated as a complete test

Answer: A

Differentiating the Cauchy-Riemann equations and adding eliminates \( v \):

\[ u_{xx} + u_{yy} \;=\; v_{yx} - v_{xy} \;=\; 0. \]

So both the real and the imaginary parts are harmonic, and this is what links complex analysis to two-dimensional potential problems in fluid flow and electrostatics

Answer: D

They are conjugate harmonic functions, and their level curves cross at right angles, since the Cauchy-Riemann equations make the two gradients perpendicular. Given one, the other is determined up to an additive constant, and the Milne-Thomson method recovers \( f(z) \) from either

Answer: C

Since \( e^{z + 2\pi i} = e^{z}e^{2\pi i} \) and \( e^{2\pi i} = 1 \),

\[ e^{z + 2\pi i} \;=\; e^{z}. \]

So the exponential is periodic with period \( 2\pi i \), a purely imaginary period. This is the reason the complex logarithm is many-valued: the exponential is not one to one

Answer: B

Writing \( z = re^{i\theta} \),

\[ \log z \;=\; \ln r + i(\theta + 2n\pi), \qquad n \in \mathbb{Z}, \]

so every choice of \( n \) gives a value. Restricting \( \theta \) to a range of width \( 2\pi \) picks the principal value, and the branch cut is where that restriction is imposed

Answer: A

On the imaginary axis, with \( z = iy \),

\[ \sin(iy) \;=\; i\sinh y, \]

whose modulus grows without bound as \( y \) does. The familiar bound of one holds only on the real axis; in the complex plane the trigonometric and hyperbolic functions are the same family seen in two directions

Answer: D

An analytic function with \( f'(z) \neq 0 \) rotates every direction through the point by \( \arg f'(z) \) and scales it by \( |f'(z)| \), so angles between curves survive unchanged. Lengths and areas are scaled rather than preserved, and directions are all rotated together, which is precisely why the angle between them is kept