Complex Variables Questions and Answers

Answer: C

The first and third hold. The second reverses the implication: the integral of \( 1/(z^{2}-1) \) round \( |z| = 2 \) vanishes although two poles lie inside, the residues merely cancelling. A vanishing integral is therefore evidence of nothing on its own

Answer: B

The exponential satisfies the Cauchy-Riemann equations at every point and is its own derivative, so it is entire. The conjugate fails them everywhere, the reciprocal fails at the origin, and the modulus is real-valued and non-constant, which no analytic function can be

Answer: A

The maximum modulus principle forbids an interior maximum unless the function is constant: the largest modulus is taken on the boundary. The Taylor expansion, the vanishing contour integral on a simply connected region, and the harmonicity of the real part are all genuine consequences of analyticity

Answer: D

Harmonicity of each part is necessary but not sufficient: the two must be conjugate, that is linked by the Cauchy-Riemann equations. Taking \( u = x \) with \( v = y \) gives two harmonic functions, yet \( x + iy \) written that way is \( \bar{z} \)'s cousin and fails the equations, so the pairing matters as much as the parts

Answer: C

For a pole of order two the recipe is

\[ \mathrm{Res}_{z=a} f \;=\; \lim_{z \to a} \frac{d}{dz}\left[(z-a)^{2}f(z)\right], \]

so the squared factor removes the pole and the differentiation extracts the coefficient \( a_{-1} \) rather than \( a_{-2} \). A double pole does have a residue, and it may well be complex

Answer: B

The real part of cos(ix) is cosh x.

From the exponential form of the cosine:

\[ \begin{aligned} \cos z &= \frac{e^{iz} + e^{-iz}}{2} \\ \cos(ix) &= \frac{e^{-x} + e^{x}}{2} = \cosh x \end{aligned} \]

For real x this is already real, so the real part is cosh x itself and the imaginary part is zero. sinh x is the corresponding result for \(-i\sin(ix)\), and cos x and sin x would be the answers only if the argument were real.

Answer: D

For the imaginary cube roots of unity \(1 + \omega + \omega^2 = 0\), so \(1 + \omega^2 = -\omega\). Then

since \(\omega^3 = 1\). The options 8, 8 omega and 8 omega squared have the wrong sign or an unreduced power of omega: they come from treating the cube of -2 omega as positive or forgetting that omega cubed equals 1.

Answer: A

The value is 2.

Expand the cube using \(i^{2} = -1\) and \(i^{3} = -i\), then divide:

\[ \begin{aligned} (i + 1)^{3} &= i^{3} + 3i^{2} + 3i + 1 \\ &= -i - 3 + 3i + 1 = 2i - 2 \\ \frac{2i - 2}{i - 1} &= \frac{2(i - 1)}{i - 1} = 2 \end{aligned} \]

A quicker route: \((1 + i)^{2} = 2i\), so \((1 + i)^{3} = 2i(1 + i) = 2i - 2\). The answer is the real number 2. −2 comes from a sign slip in \(i^{3}\) or in the denominator, and \((i + 1)\) and \((i - 1)\) leave the result complex, which it is not.

Answer: B

The required value is k = 1.

Substitute z = x + iy:

\[ \begin{aligned} F &= i(x + iy) + kx + iy \\ &= (kx - y) + i(x + y) \end{aligned} \]

So u = kx - y and v = x + y. The Cauchy-Riemann equations need \(u_{x} = v_{y}\), which gives k = 1, and \(u_{y} = -v_{x}\), which gives -1 = -1 and holds for any k.

k = 0 or -1 breaks the first equation, and k = y is not a constant; with k = 1, F(z) = (1 + i)z, which is plainly analytic.

Answer: A

The integral equals \((-1 + i)\pi\).

The denominator factorises, as substituting \(z = i\) suggests (it gives \(2 + 3i - 7 - 3i + 5 = 0\)):

\[ \begin{aligned} &2z^{4} - 3z^{3} + 7z^{2} - 3z + 5 \\ &= (z^{2} + 1)(2z^{2} - 3z + 5) \end{aligned} \]

So \(z = i\) is a simple pole, and its residue is

\[ \begin{aligned} \text{Res} &= \frac{6i}{(2i)(2i^{2} - 3i + 5)} = \frac{3}{3 - 3i} \\ &= \frac{1}{1 - i} = \frac{1 + i}{2} \end{aligned} \]

By the residue theorem the integral is \(2\pi i \times \frac{1 + i}{2} = \pi(i - 1)\).

\((1 + i)\pi\) is the residue times \(2\pi\) without the factor \(i\), and the other two options do not come from this residue.