Exercise 2

Let \(g(z) = f(z)f(-z)f(iz)f(-iz)\). For any two edges \(e_1, e_2\) along \(\partial\Omega\) it follows from \(g(z) = g(iz)\) that \[ \sup_{z\in e_1} \left\lvert g(z)\right\rvert = \sup_{z\in e_2} \left\lvert g(z)\right\rvert.\] By maximum modulus principle \(g\) attains its maximum on \(\partial\Omega\), so we have \[\begin{align*} \left\lvert g(0)\right\rvert &\leq \sup_{z\in\partial\Omega} \left\lvert g(z)\right\rvert\\ &= \sup_{z\in e} \left\lvert g(z)\right\rvert \\ &= \sup_{z\in e} \left\lvert f(z)\right\rvert \left\lvert f(iz)\right\rvert\left\lvert f(-z)\right\rvert\left\lvert f(-iz)\right\rvert \\ &\leq \sup_{z\in e} \left\lvert f(z)\right\rvert \left(\sup_{z\in \partial\Omega} \left\lvert f(z)\right\rvert\right)^3 \\ &= \sup_{z\in e} \left\lvert f(z)\right\rvert \left(\sup_{z\in \overline{\Omega}} \left\lvert f(z)\right\rvert\right)^3 \\ &= m M^3 \end{align*}\] as \(\left\lvert g(0)\right\rvert = \left\lvert f(0)\right\rvert^4\) the result follows immediately

Exercise 4

Suppose there is \(v\) such that \(f = u + iv\) is holomorphic on the punctured plane. For any open disk \(D\), \(f(D\setminus\left\{0\right\})\) the image of the punctured disk is a neighbourhood of \(\infty\), in particular it contains \(\mathbb{C}\setminus \overline{\mathbb{D}_r}\) for some positive radius \(r\). As a result \(\Re(f(D\setminus\left\{0\right\})) = u(D\setminus\left\{0\right\}) = \mathbb{R}\). Let radius of \(D\) tend to zero and \(u\) sending \((D\setminus\left\{0\right\})\) to the entire real line will contradict \(\lim_{z\to 0} u(z) = +\infty\).

An example is \(u(z) = - \log(\left\lvert z\right\rvert)\).

Exercise 6

Let \(u\) be harmonic such that \(u^3\) is also harmonic, then \[\begin{align*} 0 &= \Delta u^3 \\ \frac{\partial}{\partial x} u^3 &= 3u^2 u_x \\ \frac{\partial^2}{\partial x^2} u^3 &= 3u^2 u_{xx} + 6u_x^2 u \\ \frac{\partial^2}{\partial y^2} u^3 &= 3u^2 u_{yy} + 6u_y^2 u \\ 0 &= 3u^2 \underbrace{(u_{xx} + u_{yy})}_{0\text{ as \(u\) is harmonic}} + 6u (u_x^2 + u_y^2) \end{align*}\] we have

• \(u\equiv 0\), or
• \(u_x \equiv u_y \equiv 0\) which means that \(u\) is constant.

Exercise 8

Easy to check that each coordinate is holomorphic. To see that \(\tau\) is injective suppose \((t^2,t^3) = (s^2,s^3)\), then \[ t^2\cdot t = t^3 = s^2 \cdot s = t^2\cdot s \implies t = s. \] To show \(\tau\) is proper, as \(\mathbb{C}^2\) is locally compact and \(\mathbb{C}\) is Hausdorff, we show that the pullback of any point in \(\mathbb{C}^2\) by \(\tau\) is compact. For \((z_1,z_2)\in \mathbb{C}^2\), \[ \tau^{-1}(\left\{(z_1,z_2)\right\}) = \left\{t: t^2=z_1 \text{ and } t^3 = z_2\right\} = \begin{cases} \left\{\frac{z_2}{z_1}\right\} &\text{if \((z_1,z_2)\) lies in the image} \\ \emptyset &\text{otherwise} \end{cases} \] which in both cases results in a compact subset of \(\mathbb{C}\). It is easy to see that \(\tau\) restricted to \(\mathbb{C}^*\) does not map anything to \(0\).

Now suppose \(\tau(\mathbb{C})\) is a submanifold of \(\mathbb{C}^2\), then by definition of submanifold, near \((0,0)\) \(\tau(\mathbb{C})\) looks like \(\begin{pmatrix} f\\g\end{pmatrix}(\mathbb{D})\) where \(f,g\) are holomorphic functions on the unit disk and \(\begin{pmatrix} f\\g\end{pmatrix}(0) = \begin{pmatrix}0\\0\end{pmatrix}\) (without loss of generality assume everything is normalised nicely). Now for all \(z\in\mathbb{D}\), \[ \begin{pmatrix} f\\g\end{pmatrix}(z) = \begin{pmatrix}t^2\\t^3\end{pmatrix} \] for some \(t\). This implies that \(f^3 = g^2\). Now express \(f\) and \(g\) as Taylor series and solving for \(f^3 = g^2\) gives us that both Taylor series have 0 as their linear term. This implies that \(\begin{pmatrix}f'\\g'\end{pmatrix}(0) = \begin{pmatrix}0\\0\end{pmatrix}\). We have shown that rank of Jacobian at \(0\) is \(0\), but earlier we showed that \(\tau(\mathbb{C}^*)\) is a submanifold of \(\mathbb{C}^2\setminus\left\{0\right\}\) (with dimension 1) so \(\tau(\mathbb{C})\) cannot be a submanifold of \(\mathbb{C}^2\).