Analytic Functions and Cauchy-Riemann Equations Questions and Answers

Answer: C

Such a function exists for the zeros \(f\left(1 - \frac{1}{n}\right) = 0\).

By the identity theorem, the zeros of a non-constant analytic function on \(\mathbb{D}\) cannot accumulate at a point inside \(\mathbb{D}\). The points \(\frac{i}{n}\) and \(\frac{1}{n}\) accumulate at 0, and \(\frac{1}{2} - \frac{1}{n}\) accumulate at \(\frac{1}{2}\), both inside the disc, so those options force \(f \equiv 0\).

The points \(1 - \frac{1}{n}\) accumulate only at 1, on the boundary, which is allowed. An example is

\[ f(z) = \sin\left(\frac{\pi}{1 - z}\right) \]

which is analytic on \(\mathbb{D}\) and vanishes where \(\frac{1}{1-z} = n\), that is at \(z = 1 - \frac{1}{n}\).

Answer: A

The principal value is \(e^{2}\).

A complex power is defined through the principal logarithm: \(a^{b} = e^{b\,\mathrm{Log}\,a}\). For \(a = -1\), \(\mathrm{Log}(-1) = \ln 1 + i\pi = i\pi\), since the principal argument of \(-1\) is \(\pi\).

\[ \begin{aligned} (-1)^{-2i/\pi} &= e^{(-2i/\pi)(i\pi)} \\ &= e^{-2i^{2}} = e^{2} \end{aligned} \]

The result is real because \(i \cdot i = -1\) cancels the minus sign. \(e^{-2}\) comes from dropping that sign, and \(e^{2i}\) or \(e^{-2i}\) from forgetting to multiply the two factors of \(i\) together.

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.

Answer: C

The correct statement is Every entire function \(f\) satisfying (P) has no zeroes in \(\mathbb{C}\).

Let \(g(z) = f(z)\overline{f(\overline{z})}\). The function \(z \mapsto \overline{f(\overline{z})}\) is entire, so \(g\) is entire, and for real \(z\), \(g(z) = |f(z)|^{2} = 1\). An entire function equal to 1 on the real line is identically 1, so \(f(z)\overline{f(\overline{z})} = 1\) everywhere and \(f\) can never vanish.

This rules out the option with infinitely many zeroes. \(f(z) = e^{iz}\) satisfies (P) and is not constant, so 'every such function is constant' is false. A non-constant polynomial has \(|p(x)| \to \infty\) as \(x \to \infty\) along the real axis, so no non-constant polynomial satisfies (P).

Answer: C

The function is \(f(z) = e^{z}\).

\(e^{z}\) is entire, and its Taylor polynomials \(\sum_{k \le n} z^{k}/k!\) converge to it uniformly on the compact set \(S^{1}\).

For a uniform limit \(f\) of polynomials on \(S^{1}\), \(\oint_{S^{1}} f(z)\,z^{m}\,dz = 0\) for every \(m \ge 0\), since each polynomial has this property. On \(S^{1}\), \(\overline{z} = 1/z\), and \(\oint \frac{1}{z}\,dz = 2\pi i \ne 0\). \(\text{Re}(z) = \frac{1}{2}(z + 1/z)\) and \(|z + 1|^{2} = 2 + z + 1/z\) also contain the \(1/z\) term, so their integrals against \(z^{0}\) are non-zero. Hence none of them is a uniform limit of polynomials on \(S^{1}\).

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.

Answer: B

f is differentiable at \(z = 5\).

Write \(|z|^2 = z\bar{z}\), so \(f = z\bar{z} - 5\bar{z} + 2\). A function is complex differentiable at a point exactly when its Wirtinger derivative with respect to \(\bar{z}\) vanishes there (the Cauchy-Riemann equations):

\[ \frac{\partial f}{\partial \bar{z}} = z - 5 = 0 \iff z = 5 \]

So f is differentiable at 5 and nowhere else. At \(5i\), \(-5i\) and \(-5\) the derivative \(z - 5\) is nonzero, so the Cauchy-Riemann equations fail.

Answer: A

The modulus is \(1/\sqrt{2}\) and the argument is \(5\pi/4\), option \(\frac{1}{\sqrt{2}}, \frac{5\pi}{4}\).

Multiply numerator and denominator by the conjugate \(-1 - i\):

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

Its modulus is \(\sqrt{1/4 + 1/4} = 1/\sqrt{2}\). The point \((-1/2, -1/2)\) lies in the third quadrant on the line \(y = x\), so its argument is \(\pi + \pi/4 = 5\pi/4\) (equivalently \(-3\pi/4\)).

\(3\pi/4\) is the argument of \(-1 + i\) itself, \(7\pi/4\) points into the fourth quadrant, and \(\sqrt{2}\) is the modulus of \(-1 + i\), not of its reciprocal.

Answer: B

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

Use the addition formula, which holds for complex arguments:

\[ \sin(x + iy) = \sin x \cos(iy) + \cos x \sin(iy) \]

From the exponential definitions, \(\cos(iy) = \dfrac{e^{-y} + e^{y}}{2} = \cosh y\) and \(\sin(iy) = \dfrac{e^{-y} - e^{y}}{2i} = i\sinh y\). Substituting gives \(\sin x \cosh y + i\cos x \sinh y\).

\(\sin x \cos y + i\cos x \sin y\) treats \(y\) as a real angle, forgetting that the imaginary argument turns cos and sin into cosh and sinh. The other two options pair the functions wrongly; at \(y = 0\) they do not even reduce to \(\sin x\).

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.