Two harmonic functions \( u \) and \( v \) related by the Cauchy-Riemann equations are described as
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
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What is the period of \( e^{z} \) in the complex plane?
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
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How many values does \( \log z \) take for a given non-zero \( z \)?
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
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Can \( \sin z \) exceed one in modulus?
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
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A mapping is described as conformal at a point when it preserves
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
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Where does \( w = z^{2} \) fail to be conformal?
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
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The bilinear transformation \( w = \dfrac{az + b}{cz + d} \) with \( ad - bc \neq 0 \) carries circles and lines to
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
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How many fixed points does a bilinear transformation have, other than the identity?
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
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Under \( w = 1/z \), the image of the circle \( |z| = 2 \) is
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
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Cauchy's integral theorem states that \( \displaystyle\oint_{C} f(z)\,dz \) vanishes whenever \( f \) is
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
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