Your Answer : (Not Answered)

Correct Answer : D

The value is 1/24.

The integral divided by \(2\pi i\) is the residue at 0, the coefficient of \(1/z\) in the Laurent series:

\[ \begin{aligned} z^{3}\cos\tfrac{1}{z} &= z^{3}\left(1 - \tfrac{1}{2z^{2}} + \tfrac{1}{24z^{4}} - \cdots\right) \\ &= z^{3} - \tfrac{z}{2} + \tfrac{1}{24z} - \cdots \end{aligned} \]

The coefficient of \(1/z\) is 1/24. The value 0 would hold if the function were analytic inside the circle, but it has an essential singularity at 0. -1/6 is the coefficient that \(\sin(1/z)\) would give.

Your Answer : (Not Answered)

Correct Answer : A

The answer is \(\mathrm{I}_{1} = \mathrm{I}_{2}\).

Both curves go once anticlockwise around the origin: \(\nu_{1}\) is the unit circle and \(\nu_{2}\) the circle of radius 10. They are homotopic in \(\mathbf{C} - \{0\}\), through the circles of radius \(r\) between 1 and 10, and \(f\) is analytic on the annulus between them. By Cauchy's theorem (the integral over the boundary of the annulus is zero), the two integrals are equal.

The radius does not scale the integral. For \(f(z) = 1/z\), both integrals equal \(2\pi i\), which rules out \(\mathrm{I}_{1} = 10\,\mathrm{I}_{2}\) and \(\mathrm{I}_{1} = \mathrm{I}_{2}/10\), and the relation is a definite one, not 'no relation'.

Your Answer : (Not Answered)

Correct Answer : A

The set is \(\{z : 0 \lt \mathrm{Re}(z) \le \pi/2\}\).

With \(z = x + iy\),

\[ \sin z = \sin x \cosh y + i \cos x \sinh y. \]

For \(0 \lt x \lt \pi/2\) the real part \(\sin x \cosh y\) is positive, and the half-strip maps one-to-one onto the right half plane except the ray \([1, \infty)\). That ray is the image of the edge \(x = \pi/2\), where \(\sin z = \cosh y \ge 1\). Including the edge \(\mathrm{Re}(z) = \pi/2\) therefore fills the right half plane exactly.

The full strip \(-\pi/2 \le \mathrm{Re}(z) \le \pi/2\) also covers the left half plane, and the horizontal strips in Im(z) contain points with \(\sin x \lt 0\).

Your Answer : (Not Answered)

Correct Answer : B

The function \(\cos z\) is onto.

For any \(w \in \mathbf{C}\), solving \(\cos z = w\) means \(e^{iz} + e^{-iz} = 2w\), that is, \(u^{2} - 2wu + 1 = 0\) with \(u = e^{iz}\). This quadratic always has a root \(u\), and \(u \neq 0\) because the product of the roots is 1, so \(u = e^{iz}\) can be solved for \(z\). Hence every complex number is a value of \(\cos\).

It is not one-one, since \(\cos(z + 2\pi) = \cos z\), so it is not bijective either. It is not bounded: \(\cos(iy) = \cosh y \to \infty\). By Liouville's theorem no non-constant entire function is bounded.

Your Answer : (Not Answered)

Correct Answer : C

The true relation is \(|x| + |y| \ge |z|\).

\[ \begin{aligned} (|x| + |y|)^{2} &= x^{2} + y^{2} + 2|x||y| \\ &\ge x^{2} + y^{2} = |z|^{2} \end{aligned} \]

Taking square roots gives \(|x| + |y| \ge |z|\), with equality only when \(x = 0\) or \(y = 0\). This is just the triangle inequality for \(z = x + iy\).

Equality (A) fails for \(z = 1 + i\), where \(2 \gt \sqrt{2}\). The \(\sqrt{2}\) version (B) is the upper bound \(|x| + |y| \le \sqrt{2}|z|\) turned into an equation, which holds only when \(|x| = |y|\), and D reverses the inequality.

Your Answer : (Not Answered)

Correct Answer : D

The conclusion is \(f'' = cf'\) for some constant \(c\).

Reading the condition as \(|f'(z)| \lt |f(z)|\): if \(f(z_{0}) = 0\) the inequality would say \(|f'(z_{0})| \lt 0\), impossible, so \(f\) never vanishes. Then \(g = f'/f\) is entire and \(|g| \lt 1\), so by Liouville's theorem \(g\) is a constant \(c\). Thus \(f' = cf\), and differentiating, \(f'' = cf'\).

In fact \(f = Ae^{cz}\) with \(c \ne 0\) (as \(f\) is non-constant). So \(1/f\) is analytic, \(f\) is not a polynomial, and \(f' = cAe^{cz}\) never vanishes, which rules out the other options.

Your Answer : (Not Answered)

Correct Answer : A

The answer is a = 0 and b = 0.

\(f\) is entire and vanishes on the disc D, a set with limit points, so by the identity theorem \(f \equiv 0\) on \(\mathbf{C}\). Multiplying \(ae^{z} + be^{-z} = 0\) by \(e^{z}\) gives \(ae^{2z} = -b\) for all \(z\).

If \(a \ne 0\), then \(e^{2z}\) would be the constant \(-b/a\), which is false. So \(a = 0\), and then \(b e^{-z} = 0\) forces \(b = 0\). Every option with a 1 gives a function such as \(e^{z}\) or \(e^{z} + e^{-z} = 2\cosh z\), which is not identically zero on any disc.

Your Answer : (Not Answered)

Correct Answer : A

The answer is only (a) is true.

(a) is Liouville's theorem for harmonic functions: a harmonic function on the whole plane that is bounded below (here by 0) is constant. One proof applies Harnack's inequality, or writes \(u = \mathrm{Re} f\) with \(f\) entire and applies Liouville to \(e^{-f}\), which is bounded.

(b) is false because \(\mathbf{C} - \{0\}\) is not simply connected. The function \(\log|z|\) is harmonic there, but its harmonic conjugate would have to be \(\arg z\), which cannot be defined continuously around the origin.

Your Answer : (Not Answered)

Correct Answer : A

The impossible case is \(f(D) = \{z : |z| \le 1\}\).

By the open mapping theorem, a non-constant analytic function on a domain maps it onto an open set, and a constant function maps it to a single point. The closed disc is neither open nor a single point, so it cannot be \(f(D)\).

The others are possible: \(f(z) = z\) gives \(f(D) = D\); \(f(z) = z/2\) gives the disc of radius 1/2; and any real constant function gives \(f(D) \subseteq \mathbf{R}\).

Your Answer : (Not Answered)

Correct Answer : B

\(f\) is nowhere analytic.

Here \(u = x^{2} + y^{2}\) and \(v = 2xy\). The Cauchy-Riemann equations need \(u_{x} = v_{y}\) and \(u_{y} = -v_{x}\):

\[ \begin{aligned} u_{x} &= 2x = v_{y} \\ u_{y} &= 2y,\quad -v_{x} = -2y \end{aligned} \]

The second holds only when \(y = 0\). So \(f\) is complex differentiable exactly on the real axis. Analyticity at a point needs differentiability on a whole neighbourhood, and every disc around a real point contains points with \(y \ne 0\), so \(f\) is analytic nowhere.

That is the trap in option D: differentiable on the real axis is not analytic there. 'Everywhere' and 'only at 0' are both ruled out by the same computation.

Your Answer : (Not Answered)

Correct Answer : A

The set is the line \(x = \log(2019)\).

For \(z = x + iy\), \(e^{z} = e^{x}e^{iy}\) and \(|e^{iy}| = 1\), so \(|e^{z}| = e^{x}\). The condition \(e^{x} = 2019\) gives \(x = \log 2019\) with \(y\) arbitrary: a vertical line in the plane.

The modulus does not depend on \(y\) at all, so \(y = 2019\) is wrong, and the set is neither a circle nor countable (it contains a point for every real \(y\)). A circle would come from \(|z| = \text{const}\), not \(|e^{z}| = \text{const}\).

Your Answer : (Not Answered)

Correct Answer : D

The omitted value is 3/5.

Solve \(w = \dfrac{3z + 4}{5z + 6}\) for \(z\): \(5wz + 6w = 3z + 4\), so

\[ z = \frac{4 - 6w}{5w - 3} \]

This gives a valid \(z\) for every \(w\) except \(w = 3/5\). That value is the limit of \(f(z)\) as \(z \to \infty\), the ratio of the leading coefficients, and since \(\infty\) is not in the domain it is never taken.

\(-6/5\) is the excluded point of the domain, not a missing value of the range. 5/3 and \(-4/3\) are both taken, at \(z = -9/4\) and \(z = -36/29\) respectively, as the formula for \(z\) shows.

Your Answer : (Not Answered)

Correct Answer : D

The radius of convergence is \( \frac{1}{4} \).

By the ratio test, with \( a_{n} = \frac{(2n)!}{(n!)^{2}} \):

\[ \begin{aligned} \frac{a_{n+1}}{a_{n}} &= \frac{(2n+2)(2n+1)}{(n+1)^{2}} \\ &\to 4 \text{ as } n \to \infty \end{aligned} \]

The series converges when \( 4|z - 3i| \lt 1 \), that is within a distance \( R = 1/4 \) of the centre \( 3i \). The centre does not affect the radius.

Infinity would need the ratio to tend to 0, as for \( \sum z^{n}/n! \). \( 1/2 \) would be the radius if the series were in powers of \( (z - 3i)^{2} \), and 2 is that value inverted; neither fits a series in \( (z - 3i)^{n} \).

Your Answer : (Not Answered)

Correct Answer : A

The collinear points are \(1 + 2i, 2 + 5i, 4 + 11i\).

Three points \(z_{1}, z_{2}, z_{3}\) are collinear when \(z_{2} - z_{1}\) and \(z_{3} - z_{1}\) are real multiples of each other. Here

\[ \begin{aligned} z_{2} - z_{1} &= 1 + 3i \\ z_{3} - z_{1} &= 3 + 9i = 3(1 + 3i) \end{aligned} \]

so the points lie on the line \(y = 3x - 1\).

The other sets fail the test. For \(1 - i, 2 + i, 1 + i\) the differences are \(1 + 2i\) and \(2i\); for \(i, -1 + 2i, 3 + 4i\) they are \(-1 + i\) and \(3 + 3i\); for \(0, -3 + i, 7 + 8i\) they are \(-3 + i\) and \(7 + 8i\). In each case the ratio is not real.

Your Answer : (Not Answered)

Correct Answer : C

The coefficient of \(\frac{1}{z-1}\) is 0.

Put \(w = z - 1\), so \(z = w + 1\) and the region is \(|w| \gt 1\). There \(|1/w| \lt 1\), so expand in powers of \(1/w\):

\[ \begin{aligned} f &= \frac{1}{w(w+1)} = \frac{1}{w^{2}}\cdot\frac{1}{1 + 1/w} \\ &= \frac{1}{w^{2}}\left(1 - \frac{1}{w} + \frac{1}{w^{2}} - \cdots\right) \\ &= \frac{1}{w^{2}} - \frac{1}{w^{3}} + \frac{1}{w^{4}} - \cdots \end{aligned} \]

Every term has power \(-2\) or lower, so there is no \(1/w\) term. The value 1 is the residue at \(z = 1\), the coefficient in the other expansion, valid for \(0 \lt |z-1| \lt 1\), where \(\frac{1}{w(1+w)} = \frac{1}{w}(1 - w + \cdots)\); that is the wrong region here, and \(-1\) and \(-2\) arise from neither expansion.

Your Answer : (Not Answered)

Correct Answer : C

Both S1 and S2 are TRUE.

S1: \(g = f^{2}\) is analytic and not identically zero (otherwise f would be 0, a constant). Its zeros are isolated. Near any point where \(g \neq 0\), the continuous function f is a continuous branch of \(\sqrt{g}\), which is analytic. At an isolated zero, f is continuous and analytic in a punctured neighbourhood, so by Riemann's removable singularity theorem it is analytic there too.

S2: where \(f^{2} \neq 0\), \(f = f^{3}/f^{2}\) is a quotient of analytic functions, hence analytic. At a zero of \(f^{2}\), f is 0 and \(|f| = |f^{2}|^{1/2}\) tends to 0, so f is bounded and continuous there, and the singularity of \(f^{3}/f^{2}\) is removable.

Your Answer : (Not Answered)

Correct Answer : C

The value is \(-e^{3\pi} + e^{-3\pi}\).

Recognise u as the real part of a product. With z = x + iy:

\[ \begin{aligned} e^{2z} &= e^{2x}(\cos 2y + i\sin 2y) \\ \sin 3z &= \sin 3x\cosh 3y + i\cos 3x\sinh 3y \end{aligned} \]

The real part of \(e^{2z}\sin 3z\) is exactly the given u. So f(z) = \(e^{2z}\sin 3z\) + ic for a real constant c. At z = 0 the product is 0, so v(0, 0) = c = 2:

\[ \begin{aligned} f(i\pi) &= e^{2\pi i}\sin(3\pi i) + 2i \\ &= i\sinh 3\pi + 2i \\ 4 + 2if(i\pi) &= 4 - 2\sinh 3\pi - 4 \\ &= -(e^{3\pi} - e^{-3\pi}) \end{aligned} \]

That is \(-e^{3\pi} + e^{-3\pi}\). The option \(e^{3\pi} - e^{-3\pi}\) has the sign lost, and the two with a plus between the exponentials would come from cosh rather than sinh.

Your Answer : (Not Answered)

Correct Answer : C

One value is \(a = 3 + i\).

The three vertices lie on the unit circle at angles \(90^{\circ}\), \(225^{\circ}\) and \(315^{\circ}\), so the triangle encloses the origin but not \(z = 2\). The term \(1/(z - 2)^{4}\) is analytic inside \(C\) and the constant 4 integrates to zero, so only the pole at 0 contributes:

\[ \begin{aligned} -(a - 2)^{2}\cdot 2\pi i &= 4\pi \\ (a - 2)^{2} &= \frac{4\pi}{-2\pi i} = 2i \\ a - 2 &= \pm(1 + i) \end{aligned} \]

since \((1 + i)^{2} = 2i\). So \(a = 3 + i\) or \(a = 1 - i\), and only \(3 + i\) is offered. \(1 + i\), \(2 + i\) and \(4 + i\) give \((a-2)^{2} = -2i\), \(-1\) and \(3 + 4i\), none equal to \(2i\).

Your Answer : (Not Answered)

Correct Answer : D

The supremum is \(\dfrac{3}{5}\).

Every bijective analytic map of the upper half-plane onto the unit disc sending i to 0 has the form

\[ f(z) = e^{i\theta}\,\frac{z - i}{z + i} \]

for some real \(\theta\),

because any two such maps differ by an automorphism of the disc fixing 0, which is a rotation. So for every f in S,

\[ |f(4i)| = \left|\frac{4i - i}{4i + i}\right| = \frac{3}{5} \]

The value is the same for all of them, so the supremum is 3/5. Zero would need \(f(4i) = f(i)\), impossible for an injective map, and 1/4 and 1/2 do not come from the Cayley form.

Your Answer : (Not Answered)

Correct Answer : A

The value is \(\dfrac{1}{7}\).

Since \(\tfrac{1}{2} \mapsto \infty\), write \(T(z) = (az + b)/(2z - 1)\). Then \(T(0) = -b = -3\) gives b = 3, and \(T(1) = a + 3 = 2\) gives a = -1, so \(T(z) = (3 - z)/(2z - 1)\).

A circle maps to a line exactly when it passes through the point sent to \(\infty\), here 1/2; the circle centred at 1 must therefore have radius \(k = \tfrac{1}{2}\). It also passes through 3/2 and \(1 + \tfrac{i}{2}\), whose images are

\[ \begin{aligned} T(\tfrac{3}{2}) &= \tfrac{3}{4} \\ T(1 + \tfrac{i}{2}) &= \tfrac{3}{4} - \tfrac{5}{4}i \end{aligned} \]

so the image is the line \(x = 3/4\): \(\alpha = 1\), \(\beta = 0\), \(\gamma = -3/4\). Then

\[ \begin{aligned} \frac{2k(\alpha + \beta) + \gamma}{\alpha + \beta - 2k\gamma} &= \frac{1 - \frac{3}{4}}{1 + \frac{3}{4}} \\ &= \frac{1}{7} \end{aligned} \]

The ratio does not depend on how the line's equation is scaled.