Complex Variables Questions and Answers

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.

Answer: A

The value of A is 2/5.

The integrand has a single pole where 5z - 4 = 0, at z = 4/5, which lies inside the unit circle. Write it as

\[ \frac{1}{5z-4} = \frac{1}{5}\cdot\frac{1}{z - 4/5} \]

so the residue at the pole is 1/5. By the residue theorem,

\[ \begin{aligned} \oint_{C}\frac{dz}{5z-4} &= 2\pi i \times \frac{1}{5} \\ &= \frac{2}{5}\pi i \end{aligned} \]

Hence A = 2/5. Forgetting the factor 1/5 gives A = 2; 4/5 is the location of the pole, not the residue; 1/2 has no route from the integral.

Answer: D

The function is \(f(z) = \cosh z\).

Expand \(\cosh z\) with \(z = x + iy\):

\[ \cosh(x + iy) = \cosh x\cos y + i\sinh x\sin y \]

Its real part is \(\cosh x\cos y\), as required, and its imaginary part \(\sinh x\sin y\) is zero when \(y = 0\). Since \(\cosh z\) is analytic and an analytic function is fixed by its real part up to an imaginary constant, which the condition at \(y = 0\) sets to zero, this is the function.

\(\cosh x\,e^{-iy}\) has the right real part but is not analytic (it fails the Cauchy-Riemann equations). \(\cosh z\cos y\) mixes z with y and is not analytic, and \(\cosh z\,e^{z}\) has a different real part.

Answer: A

The correct statement is \(w\) is analytic everywhere except at \(z = 0\).

Write \(w = u + iv\) with \(u = \frac{1}{2}\log_{e}(x^{2} + y^{2})\) and \(v = \tan^{-1}(y/x)\). Then

\[ \begin{aligned} u_{x} &= \frac{x}{x^{2} + y^{2}} = v_{y} \\ u_{y} &= \frac{y}{x^{2} + y^{2}} = -v_{x} \end{aligned} \]

so the Cauchy-Riemann equations hold wherever \(x^{2} + y^{2} \ne 0\), and \(dw/dz = 1/z\) exists for every \(z \ne 0\). (Taking a single-valued branch also needs a cut from the origin, but the textbook answer, and the key, single out \(z = 0\).)

So \(w\) is not non-analytic everywhere. Its conjugate pair is \(\frac{1}{2}\log_{e}(x^{2} + y^{2})\) and \(\tan^{-1}(y/x)\), not \(\log_{e}(x^{2} - y^{2})\) and not \(\tan^{-1}(x/y)\).

Answer: D

Neither f nor g is analytic.

For \(g(z) = \bar{z} = x - iy\): u = x, v = -y, so \(u_{x} = 1\) but \(v_{y} = -1\). The Cauchy-Riemann condition \(u_{x} = v_{y}\) fails everywhere, so g is nowhere differentiable.

For \(f(z) = |z|\): f is real-valued, so v = 0 and the conditions would force \(u_{x} = u_{y} = 0\), making f constant, which it is not. (At the origin |z| is not even differentiable as a real function.) Since both fail, every option claiming that one or both are analytic is wrong.

Answer: C

The value is 7.

Split the integral into two parts. \(\cos z = 0\) only at \(z = \pm\pi/2, \pm 3\pi/2, \ldots\), all outside \(|z| = 1\) since \(\pi/2 \gt 1\). So \(e^{z}/\cos z\) is analytic inside and on \(\gamma\), and by Cauchy's theorem its integral is 0.

For f, the only singularity inside \(\gamma\) is at 0, and the residue there is the coefficient of \(1/z\), which is \(7/2\); the \(1/(8z^{2})\) term and the power series contribute nothing. By the residue theorem

\[ \begin{aligned} \oint_{\gamma} f\,dz &= 2\pi i \cdot \frac{7}{2} = 7\pi i \\ \frac{1}{\pi i}\cdot 7\pi i &= 7 \end{aligned} \]

7/2 is the residue itself without the \(2\pi i / \pi i = 2\) factor, 1/8 uses the coefficient of \(1/z^{2}\), and 0 is what the cosine term alone contributes; it would be the whole answer only if f had no \(1/z\) term.