Analytic Functions and Cauchy-Riemann Equations Questions and Answers

Answer: D

The impossible case is \(f(G) = \{z : |z| = 1\}\) (G being the unit disc).

By the open mapping theorem, a non-constant analytic function maps open sets to open sets, so \(f(G)\) is open. The unit circle contains no open disc, so it cannot be the image.

The other images are possible: \(f(z) = z + 2\) maps the disc into \(\mathrm{Re}(z) \ge 1\); \(f(z) = 2z\) maps it onto \(|z| \lt 2\); and the open set \(|z| \gt 1\) is also the image of a suitable analytic map (for instance \(e^{g(z)}\) with \(g\) mapping the disc onto the half plane \(\mathrm{Re} \gt 0\)).

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.

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.

Answer: A

The answer is \(H(\Omega)\) is an integral domain which does not satisfy the ascending chain condition on ideals.

If \(fg = 0\) on a region, the zeros of a non-zero analytic function are isolated, so one of \(f, g\) must vanish identically: \(H(\Omega)\) is an integral domain. It is not Noetherian: pick a sequence of distinct points \(a_{n}\) in \(\Omega\) without a limit point in \(\Omega\), and let \(I_{n}\) be the ideal of functions vanishing at \(a_{n}, a_{n+1}, \ldots\). By Weierstrass's theorem these ideals strictly increase forever.

For the same reason \(H(\Omega)\) is not a UFD (a function with infinitely many zeros has no finite factorisation), it is infinite dimensional (it contains all polynomials), and it has infinitely many maximal ideals (one for each point of \(\Omega\)).

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.

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.

Answer: B

The false statement is that \(u^{2}, v^{2}\) are harmonic.

For harmonic \(u\), \(\Delta(u^{2}) = 2u\Delta u + 2|\nabla u|^{2} = 2|\nabla u|^{2}\), which is zero only where \(\nabla u = 0\). So \(u^{2}\) is not harmonic unless \(u\) is constant. For \(f = z\), \(u = x\) and \(u^{2} = x^{2}\) has Laplacian 2.

The other statements hold. The real and imaginary parts of an analytic function are harmonic (A), and so is any sum of harmonic functions (C). Also \(f^{2} = u^{2} - v^{2} + 2iuv\) is analytic, so its imaginary part \(2uv\), and hence \(uv\), is harmonic (D).

Answer: D

One value of \(i^{(1+i)}\) is \(i e^{-\pi/2}\).

A complex power is defined by \(z^{w} = e^{w \log z}\), with \(\log i = i(\pi/2 + 2k\pi)\). Taking the principal branch \(k = 0\):

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

Other branches multiply this by \(e^{-2k\pi}\), so every value is \(i\) times a positive real number. \(i e^{\pi/2}\) has the wrong sign in the exponent (it would need \(k = -1/2\)), and \(e^{\pm \pi i/2} = \pm i\) has modulus 1, which none of the values has.

Answer: B

The conjugate is \(u = x + x^{2} - y^{2} + c\).

For \(f = u + iv\) to be analytic, the Cauchy-Riemann equations \(u_{x} = v_{y}\) and \(u_{y} = -v_{x}\) must hold. With \(v = y + 2xy\):

\[ \begin{aligned} u_{x} &= v_{y} = 1 + 2x \\ u &= x + x^{2} + \phi(y) \\ u_{y} &= \phi'(y) = -v_{x} = -2y \\ \phi(y) &= -y^{2} + c \end{aligned} \]

So \(u = x + x^{2} - y^{2} + c\) and \(f(z) = z + z^{2} + c\). The option \(x^{2} - y^{2} + c\) drops the \(x\) that comes from the \(y\) in \(v\); \(x + 2xy + c\) just copies the form of \(v\); and \(y^{2} - x^{2} + y + c\) has the signs reversed and fails \(u_{x} = 1 + 2x\).

Answer: C

An analytic \(f\) with \(f(D)\) inside a circle must be constant.

A circle is a curve with empty interior in C. By the open mapping theorem, a non-constant analytic function on a domain maps open sets to open sets, so \(f(D)\) would contain a disk, which a circle cannot. Hence \(f\) is constant. The same follows from the Cauchy-Riemann equations: if \(|f - c|\) is constant, so is \(f\).

The other sets have interior, so they impose nothing: \(f(z) = z\) maps the unit disk into a disk, \(f(z) = z + 2\) maps it into the half plane Re \(w \gt 0\), and every analytic function maps into C. (A half plane forces constancy only for entire functions, by Liouville's theorem; here \(f\) is defined only on a disk.)