Complex Variables Questions and Answers

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.

Answer: C

\(|f'(-1 + i)|^{2} = 8\).

For an analytic function \(f'(z) = u_{x} + i\,v_{x}\). From the given parts:

\[ \begin{aligned} u_{x} &= y^{3} - 3x^{2}y \\ v_{x} &= x^{3} - 3xy^{2} \end{aligned} \]

(The Cauchy-Riemann equation \(u_{x} = v_{y}\) checks: \(v_{y} = y^{3} - 3x^{2}y\).) At \(x = -1\), \(y = 1\):

\[ \begin{aligned} u_{x} &= 1 - 3 = -2,\quad v_{x} = -1 + 3 = 2 \\ |f'|^{2} &= (-2)^{2} + 2^{2} = 8 \end{aligned} \]

The values 1, 7 and 10 come from sign slips in these two derivatives.

Answer: A

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

At \(z = 0\) the difference quotient is \(\frac{f(h) - f(0)}{h} = |h| \to 0\), so \(f'(0) = 0\) exists. Elsewhere write \(r = \sqrt{x^{2} + y^{2}}\), \(u = xr\), \(v = yr\). Then

\[ \begin{aligned} u_{x} &= r + \tfrac{x^{2}}{r}, \quad v_{y} = r + \tfrac{y^{2}}{r} \\ u_{y} &= \tfrac{xy}{r}, \quad v_{x} = \tfrac{xy}{r} \end{aligned} \]

The Cauchy-Riemann equations \(u_{x} = v_{y}\) and \(u_{y} = -v_{x}\) need \(x^{2} = y^{2}\) and \(xy = 0\), which together force \(x = y = 0\). So \(z|z|\) is differentiable only at the origin.

\(\sin z\) and \(e^{-z^{2}}\) are entire, differentiable everywhere. \(e^{1/z}\) is differentiable at every \(z \ne 0\) but has an essential singularity at 0, so it is not even continuous there: the reverse of what is asked.

Answer: A

The correct statement is \(f\) is periodic.

\(e^{z + 2\pi i} = e^{z}e^{2\pi i} = e^{z}\), so \(e^{z}\) is periodic with the purely imaginary period \(2\pi i\), even though the real exponential \(e^{x}\) is not periodic.

Writing \(e^{z} = e^{x}(\cos y + i\sin y)\) settles the others. The modulus is \(|f| = e^{x}\), which equals 1 only on the imaginary axis. The argument is \(y + 2n\pi\), not \(y \pm n\pi\): adding an odd multiple of \(\pi\) would flip the sign of f. And since f is periodic, 'f is not periodic' is false.

Answer: B

The value is \(f(1) = 1 + 5i\).

Look for an analytic \(f'(z)\) whose real part is \(3x^{2}-4y-3y^{2}\). Since \(\text{Re}(3z^{2}) = 3x^{2}-3y^{2}\) and \(\text{Re}(4iz) = -4y\):

\[ f'(z) = 3z^{2} + 4iz + ic \]

for a real constant c, and f'(0) = 0 gives c = 0. Integrating:

\[ \begin{aligned} f(z) &= z^{3} + 2iz^{2} + K \\ f(i) &= -i - 2i + K = 0 \Rightarrow K = 3i \\ f(1) &= 1 + 2i + 3i = 1 + 5i \end{aligned} \]

Once f'(z) is fixed, f(i) = 0 fixes K, so f(1) has only one possible value. Dropping K would give 1 + 2i, which is not offered; 4 + 2i, 4 - 2i and 1 - i do not come from any f with this derivative that vanishes at i.

Answer: C

Both P and Q are true, and Q implies P.

Write \(u = x^{3}-4xy^{2}\) and \(v = 4x^{2}y-2y^{3}\). Then:

\[ \begin{aligned} u_{x} &= 3x^{2}-4y^{2}, & v_{y} &= 4x^{2}-6y^{2} \\ u_{y} &= -8xy, & v_{x} &= 8xy \end{aligned} \]

The second Cauchy-Riemann equation \(u_{y} = -v_{x}\) always holds, but \(u_{x} = v_{y}\) needs \(x^{2} = 2y^{2}\). On the x-axis (y = 0) that means x = 0, so the equations fail at every other point of the x-axis: Q is true.

Analyticity at a point needs the equations to hold throughout a neighbourhood of it. Every neighbourhood of (0, 0) contains points of the x-axis where they fail, so f is not analytic at the origin: P is true, and this argument derives P from Q, so Q implies P.