The given complex function is defined as:

\(f(z) =\ \left\{ \begin{matrix} \rm \frac{\bar{z}^2}{z}\ if \ z \ne 0,\\ \rm 0 \ \ otherwise. \end{matrix} \right.\)

We need to evaluate each statement to find the false one.

Continuous Complex Function f(z)

Let's check the continuity of \(f(z)\) everywhere. For \(z \ne 0\), the function \(f(z) = \frac{\bar{z}^2}{z}\) is a ratio of continuous functions (\(\bar{z}^2\) and \(z\)) where the denominator is non-zero, so \(f(z)\) is continuous for all \(z \ne 0\).

We need to check continuity at \(z=0\). The function is continuous at \(z=0\) if \(\lim_{z \to 0} f(z) = f(0)\). We are given \(f(0) = 0\).

Let's evaluate the limit \(\lim_{z \to 0} f(z)\) using polar coordinates. Let \(z = re^{i\theta}\). Then \(\bar{z} = re^{-i\theta}\).

\(f(z) = \frac{\bar{z}^2}{z} = \frac{(re^{-i\theta})^2}{re^{i\theta}} = \frac{r^2 e^{-2i\theta}}{re^{i\theta}} = r e^{-3i\theta}\).

As \(z \to 0\), the radial distance \(r \to 0\). The limit is:

\(\lim_{z \to 0} f(z) = \lim_{r \to 0} re^{-3i\theta}\).

Since \(|e^{-3i\theta}| = 1\), we have \(|re^{-3i\theta}| = r\). As \(r \to 0\), \(r \to 0\). So, \(\lim_{r \to 0} re^{-3i\theta} = 0\).

Since \(\lim_{z \to 0} f(z) = 0 = f(0)\), the function \(f(z)\) is continuous at \(z=0\).

Therefore, \(f(z)\) is continuous everywhere in \(\complexes\).

Statement 1: f(z) is continuous everywhere - This statement is TRUE.

Analytic Function Properties near Zero

A function is analytic in an open neighbourhood of zero if it is differentiable at every point in that neighbourhood. If it is analytic at \(z=0\), it must be differentiable at \(z=0\).

Let's check the differentiability of \(f(z)\) at \(z=0\). The derivative \(f'(0)\) exists if the limit \(\lim_{z \to 0} \frac{f(z) - f(0)}{z - 0}\) exists.

\(f'(0) = \lim_{z \to 0} \frac{f(z)}{z}\).

For \(z \ne 0\), \(\frac{f(z)}{z} = \frac{\bar{z}^2/z}{z} = \frac{\bar{z}^2}{z^2}\).

Let's check the limit along different paths approaching zero.

• Along the real axis, \(z=x\) where \(x \to 0\): \(\frac{\bar{x}^2}{x^2} = \frac{x^2}{x^2} = 1\). The limit is 1.
• Along the imaginary axis, \(z=iy\) where \(y \to 0\): \(\frac{\overline{(iy)}^2}{(iy)^2} = \frac{(-iy)^2}{(iy)^2} = \frac{-y^2}{-y^2} = 1\). The limit is 1.
• Along the line \(y=x\), \(z=x+ix=x(1+i)\) where \(x \to 0\). \(\bar{z} = x(1-i)\). \(\frac{\bar{z}^2}{z^2} = \frac{(x(1-i))^2}{(x(1+i))^2} = \frac{x^2(1-i)^2}{x^2(1+i)^2} = \frac{(1-i)^2}{(1+i)^2} = \frac{1-2i-1}{1+2i-1} = \frac{-2i}{2i} = -1\). The limit is -1.

Since the limit \(\lim_{z \to 0} \frac{f(z)}{z}\) depends on the path of approach, it does not exist. Therefore, \(f(z)\) is not differentiable at \(z=0\).

If \(f(z)\) is not differentiable at \(z=0\), it cannot be analytic in any open neighbourhood that contains \(z=0\).

Statement 2: f(z) is not analytic in any open neighbourhood of zero - This statement is TRUE.

Cauchy-Riemann Equations at Zero for zf(z)

Let \(g(z) = zf(z)\).

For \(z \ne 0\), \(g(z) = z \cdot \frac{\bar{z}^2}{z} = \bar{z}^2\).

For \(z = 0\), \(g(0) = 0 \cdot f(0) = 0 \cdot 0 = 0\).

So, \(g(z) = \left\{ \begin{matrix} \bar{z}^2 & \text{if } z \ne 0,\\ 0 & \text{if } z = 0. \end{matrix} \right.\)

We need to check if \(g(z)\) satisfies the Cauchy-Riemann (CR) equations at \(z=0\). Let \(z = x + iy\). Then \(g(z) = u(x, y) + iv(x, y)\).

For \(z \ne 0\), \(g(z) = \bar{z}^2 = (x-iy)^2 = x^2 - 2ixy + (iy)^2 = x^2 - y^2 - 2ixy\).

So, for \(z \ne 0\), \(u(x, y) = x^2 - y^2\) and \(v(x, y) = -2xy\).

At \(z=0\), \(g(0) = 0\), so \(u(0, 0) = 0\) and \(v(0, 0) = 0\).

We need to compute the partial derivatives at \((0, 0)\) using the limit definition:

\(\frac{\partial u}{\partial x}(0, 0) = \lim_{h \to 0} \frac{u(h, 0) - u(0, 0)}{h}\). For \(h \ne 0\), \(u(h, 0) = h^2 - 0^2 = h^2\). \(u(0, 0) = 0\). \(\frac{\partial u}{\partial x}(0, 0) = \lim_{h \to 0} \frac{h^2 - 0}{h} = \lim_{h \to 0} h = 0\).

\(\frac{\partial u}{\partial y}(0, 0) = \lim_{k \to 0} \frac{u(0, k) - u(0, 0)}{k}\). For \(k \ne 0\), \(u(0, k) = 0^2 - k^2 = -k^2\). \(u(0, 0) = 0\). \(\frac{\partial u}{\partial y}(0, 0) = \lim_{k \to 0} \frac{-k^2 - 0}{k} = \lim_{k \to 0} -k = 0\).

\(\frac{\partial v}{\partial x}(0, 0) = \lim_{h \to 0} \frac{v(h, 0) - v(0, 0)}{h}\). For \(h \ne 0\), \(v(h, 0) = -2(h)(0) = 0\). \(v(0, 0) = 0\). \(\frac{\partial v}{\partial x}(0, 0) = \lim_{h \to 0} \frac{0 - 0}{h} = 0\).

\(\frac{\partial v}{\partial y}(0, 0) = \lim_{k \to 0} \frac{v(0, k) - v(0, 0)}{k}\). For \(k \ne 0\), \(v(0, k) = -2(0)(k) = 0\). \(v(0, 0) = 0\). \(\frac{\partial v}{\partial y}(0, 0) = \lim_{k \to 0} \frac{0 - 0}{k} = 0\).

The Cauchy-Riemann equations at \((0, 0)\) are:

\(\frac{\partial u}{\partial x}(0, 0) = \frac{\partial v}{\partial y}(0, 0)\) \(\implies 0 = 0\) (True)

\(\frac{\partial u}{\partial y}(0, 0) = -\frac{\partial v}{\partial x}(0, 0)\) \(\implies 0 = -0 \implies 0 = 0\) (True)

The CR equations are satisfied at \(z=0\).

Statement 3: zf(z) satisfies the Cauchy-Riemann equations at zero - This statement is TRUE.

Analytic Function in Open Subset of ℂ

A function is analytic in an open subset if it is differentiable at every point in that subset.

We already showed that \(f(z)\) is not differentiable at \(z=0\). Therefore, \(f(z)\) cannot be analytic in any open subset that contains \(z=0\).

Consider an open subset \(U\) that does not contain \(z=0\). For \(f(z)\) to be analytic in \(U\), it must be differentiable at every point \(z \in U\). For \(z \ne 0\), \(f(z) = \frac{\bar{z}^2}{z}\).

Let's examine the differentiability of \(f(z) = \frac{\bar{z}^2}{z}\) for \(z \ne 0\). A function \(f(z) = u(x, y) + iv(x, y)\) is differentiable at a point if the CR equations hold at that point and the partial derivatives \(u_x, u_y, v_x, v_y\) are continuous at that point.

We used the polar form \(f(z) = re^{-3i\theta}\) earlier. Let's consider the condition for analyticity using the \(\bar{z}\) derivative. A function \(f\) is analytic if and only if \(\frac{\partial f}{\partial \bar{z}} = 0\).

For \(z \ne 0\), \(f(z) = z^{-1} \bar{z}^2\).

\(\frac{\partial f}{\partial \bar{z}} = \frac{\partial}{\partial \bar{z}} (z^{-1} \bar{z}^2)\). Since \(z\) is treated as a constant with respect to \(\bar{z}\) in this partial derivative, we get:

\(\frac{\partial f}{\partial \bar{z}} = z^{-1} \frac{\partial}{\partial \bar{z}} (\bar{z}^2) = z^{-1} (2\bar{z}) = \frac{2\bar{z}}{z}\).

For \(f(z)\) to be analytic in an open set, we need \(\frac{\partial f}{\partial \bar{z}} = 0\) at every point in that set. \(\frac{2\bar{z}}{z} = 0\) implies \(\bar{z} = 0\), which means \(z=0\).

The condition \(\frac{\partial f}{\partial \bar{z}} = 0\) is satisfied only at \(z=0\). This indicates that \(f(z)\) is not analytic in any open set containing points \(z \ne 0\).

Since \(f(z)\) is not differentiable at \(z=0\) and not analytic for \(z \ne 0\), \(f(z)\) is not analytic in any open subset of \(\complexes\).

Statement 4: f(z) is analytic in some open subset of ℂ - This statement is FALSE.

Based on our analysis, the false statement is the fourth one.