Analytic Functions and Cauchy-Riemann Equations Questions and Answers

Answer: B

The function that is not an open mapping is \(f(z) = |z|\).

By the open mapping theorem, every non-constant analytic function on a domain is open. \(e^{z^{2}}\), \(\sin z + \cos z^{2}\) and the series \(1 - z^{2}/2! + z^{4}/4! - \cdots = \cos z\) are non-constant entire functions, so they are open maps.

\(|z|\) is continuous but takes only real, non-negative values. The image of any open disc is a subset of the real line, which contains no open set of \(\mathbf{C}\), so the map is not open. (It is also not analytic, so the theorem does not apply.)

Answer: A

One value of \(i^{i}\) is \(e^{-\pi/2}\).

By definition \(i^{i} = e^{i \log i}\). With the principal branch, \(\log i = \ln|i| + i\arg i = i\pi/2\), so

\[ i^{i} = e^{i \cdot i\pi/2} = e^{-\pi/2} \approx 0.2079 \]

Other branches give \(e^{-\pi/2 - 2k\pi}\), all real and positive. The options \(e^{\pm i\pi/2}\) equal \(\pm i\), which are not values of \(i^{i}\), and \(e^{\pi/2}\) would need \(\arg i = -\pi/2\), which is the argument of \(-i\), not of \(i\).

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 \(|e^{z}| = \text{const}\).

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.

Answer: A

\(f\) must be a linear polynomial, \(f(z) = az + b\) with \(a \ne 0\).

An entire injective function cannot have an essential singularity at \(\infty\): by Casorati-Weierstrass, \(f(\{|z| \gt 1\})\) would be dense and would meet the open set \(f(\{|z| \lt 1\})\), contradicting injectivity. So \(f\) is a polynomial, and an injective polynomial must have degree 1, since a polynomial of degree \(n \ge 2\) takes most values \(n\) times.

So \(f'(z) = a \ne 0\) everywhere, ruling out (B), and \(f^{-1}(w) = (w - b)/a\) is continuous and analytic, ruling out (C) and (D).

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 \(f(D) \subseteq \mathbf{R}\).

Answer: C

The possible case is \(\bar{f} + \bar{g}\) is analytic.

\(\bar{f} + \bar{g} = \overline{f + g}\). A function and its conjugate are both analytic only when the function is constant, so this is analytic exactly when \(f + g\) is constant, for example \(g = -f + 1\) with \(f\) non-constant. So it can happen.

The others cannot. If \(f - \bar{g}\) were analytic, then \(\bar{g} = f - (f - \bar{g})\) would be analytic, forcing \(g\) constant; the same argument rules out \(f + \bar{g}\). And \(fg \equiv 0\) on a region would force \(f \equiv 0\) or \(g \equiv 0\) (the ring of analytic functions on a region has no zero divisors), contradicting non-constancy.

Answer: D

The answer is \(z = \bar{w}\) or \(z, w \in \mathbf{R}\).

If \(z + w = s\) and \(zw = p\) are both real, then \(z\) and \(w\) are the two roots of the real quadratic \(t^{2} - st + p = 0\). A real quadratic has either two real roots or a pair of complex conjugate roots. So either both are real, or \(z = \bar{w}\).

Each narrower option misses a case: \(z = 1, w = 2\) satisfies the conditions with \(z \ne w\) and \(z \ne \bar{w}\); \(z = i, w = -i\) satisfies them without being real; and \(z = w\) is not forced by either case.

Answer: C

The inequality is true if and only if \(\mathrm{Im}(z) \gt 0\).

Geometrically, \(|z - i| \lt |z + i|\) says that \(z\) is closer to \(i\) than to \(-i\). The perpendicular bisector of the segment from \(-i\) to \(i\) is the real axis, so the points closer to \(i\) form the upper half plane.

Algebraically, with \(z = x + iy\):

\[ \begin{aligned} x^{2} + (y - 1)^{2} &\lt x^{2} + (y + 1)^{2} \\ -2y &\lt 2y \iff y \gt 0 \end{aligned} \]

So it is neither never nor always true, and the real part plays no role.

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.