Exercise 1

1

We have \(\frac{\partial z}{\partial z} = \frac{\partial \overline{z}}{\partial \overline{z}} = 1, \frac{\partial z}{\partial \overline{z}} = \frac{\partial \overline{z}}{\partial z} = 0.\)

2

Without loss of generality, let \(z_0 = 0\). Then, we have that \(f(z) = u(z) + iv(z)\) for real functions \(u,v\).

\[f(x + iy) = f(0) + \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix} + o(|x + iy|)\]

Let \(\alpha = a + ib\) and \(\beta = c + id\). We get that \(\alpha(x + iy) + \beta(x - iy)\) has the form \[\begin{pmatrix}(a+c)x + (b-d)y & (-b-d)x + (a-c)y \end{pmatrix}^T = \begin{pmatrix} \frac{\partial u}{\partial x}x + \frac{\partial u}{\partial y}y & \frac{\partial v}{\partial x}x + \frac{\partial v}{\partial y}y\end{pmatrix}^T.\] This has 4 independent equations and 4 unknowns, so it can always be done.

3

Differentiating the expression above shows \(\alpha = \frac{\partial f}{\partial z}\) and \(\beta = \frac{\partial f}{\partial \overline{z}}.\)

4

This follows from Cauchy Riemann Equations. Observe that \(\frac{\partial (u + iv)}{\partial \overline{z}} = \frac{1}{2}\left( (\frac{\partial u}{\partial x} - \frac{\partial v}{\partial y}) + i(\frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}) \right).\)

5

We have that \(\frac{\partial \overline{f}}{\partial \overline{z}} = \overline{\frac{\partial f}{\partial z}}\). Thus, if \(f\) is anti-holomorphic, then \(\frac{\partial \overline{f}}{\partial \overline{z}} = \overline{0} = 0.\) Conversely, if \(\overline{f}\) is holomorphic, then, \(\frac{\partial f}{\partial \overline{z}} = \overline{\frac{\partial \overline{f}}{\partial z}} = 0.\)

Exercise 2

1

Let \(u(z) = x + iy\) be \(\mathscr{C}^2\), \[\begin{align*} \frac{\partial^2u}{\partial z\partial \overline{z}} &= \frac{\partial}{\partial z}\frac12\left( \frac{\partial u}{\partial x} + i\frac{\partial u}{\partial y} \right) \\ &= \frac12\left( \frac{\partial^2u}{\partial z\partial x} + i\frac{\partial^2u}{\partial z\partial y} \right) \\ \end{align*}\] now \[ \frac{\partial^2 u}{\partial z\partial x} = \frac12\left( \frac{\partial^2 u}{\partial x\partial x} - i\frac{\partial^2 u}{\partial y\partial x} \right) \text{ and } \frac{\partial^2 u}{\partial z\partial y} = \frac12\left( \frac{\partial^2 u}{\partial x\partial y} - i\frac{\partial^2 u}{\partial y\partial y} \right) \] so \[ \frac{\partial^2 u}{\partial z\partial\overline{z}} = \frac14\left(\frac{\partial^2 u}{\partial x\partial x} - i\frac{\partial^2 u}{\partial y\partial x} + i \frac{\partial^2 u}{\partial x\partial y} + \frac{\partial^2 u}{\partial y\partial y} \right) \] and by Clairaut’s theorem we know that \(\frac{\partial^2 u}{\partial x\partial y} = \frac{\partial^2 u}{\partial y\partial x}\) so we have \(\Delta u = 4\frac{\partial^2 }{\partial z\partial\overline{z}} u\) as desired.

\(u\) is harmonic iff \(\Delta u = 0\) iff \(\frac{\partial^2 u}{\partial z\partial\overline{z}} = 0\) iff (by Clairaut’s theorem again) \(\frac{\partial^2 u}{\partial\overline{z}\partial z} = \frac{\partial}{\partial\overline{z}} \frac{\partial u}{\partial z} = 0\) iff (by Exercise 1.4) \(\frac{\partial u}{\partial z}\) is holomorphic.

2

Let \(v = u(f(z))\), then in some open set where \(u(f(z))\) is well-defined \[ \frac{\partial v}{\partial z} = \frac{\partial u}{\partial f} f'(z) \] \(\frac{\partial u}{\partial f}\) makes sense and (by part 1) will be holomorphic as \(u\) is harmonic. And as \(f'\) is also holomorphic \(u(f(z))\) harmonic.

3

Let \(f(x + iy) = u(x,y) + iv(x,y)\), so \(\frac{\partial f}{\partial\overline{z}} = 0\), that is \[ \frac{\partial f}{\partial x} = -i \frac{\partial f}{\partial y} \] \[ u_x + i v_x = -i u_y + v_y \] so \(u_x = v_y\) and \(v_x = -u_y\). It follows that \[ u_{xx} = v_{yx} \text{ and } v_{xy} = - u_{yy} \] and by Clairaut theorem \[ \Delta \Re(f) = u_{xx} + u_{yy} = v_{yx} - v_{xy} = 0 \] similarly we can show \[ \Delta \Im(f) = v_{xx} + v_{yy} = u_{xy} - u_{yx} = 0. \]

4

Outside the zero set of \(f\), we have \[ \frac{\partial}{\partial\overline{z}}\log(f\overline{f}) = \frac{\frac{\partial f}{\partial\overline{z}}\overline{f}+ f\frac{\partial\overline{f}}{\partial\overline{z}}}{f\overline{f}} = \frac{1}{\overline{f}} \frac{\partial\overline{f}}{\partial\overline{z}} \] so \[ \frac{\partial^2 }{\partial z\partial\overline{z}}\log(f\overline{f}) = \frac{-\frac{\partial\overline{f}}{\partial z}}{(\overline{f})^2}\frac{\partial\overline{f}}{\partial\overline{z}} + \frac{1}{\overline{f}} \frac{\partial^2 \overline{f}}{\partial z\partial\overline{z}} = \frac{1}{\overline{f}} \frac{\partial^2 \overline{f}}{\partial z\partial\overline{z}} = \frac{1}{\overline{f}} \frac{\partial^2 \overline{f}}{\partial\overline{z}\partial z} = 0 \] as \(\frac{\partial\overline{f}}{\partial z} = \frac{\partial f}{\partial\overline{z}} = 0\). And therefore \[\Delta\log\left\lvert f\right\rvert = \Delta \frac12 \log\left\lvert f\right\rvert^2 = \Delta\frac12\log(f\overline{f}) = 0\] and \(\log\left\lvert f\right\rvert\) is harmonic.

5

Suppose \(\left\lvert f\right\rvert^2\) is harmonic. Let \(f(x+iy) = u(x,y) + iv(x,y)\) and \(\left\lvert f(x+iy)\right\rvert^2 = u^2(x,y) + v^2(x,y)\). \[\begin{align*} \Delta \left\lvert f^2\right\rvert &= \frac{\partial^2}{\partial x^2}(u^2 + v^2) + \frac{\partial^2}{\partial y^2}(u^2 + v^2) \\ &= \frac{\partial}{\partial x}(2uu_x + 2vv_x) + \frac{\partial}{\partial y}(2uu_y + 2vv_y) \\ &= 2(u_x)^2 + 2uu_{xx} + 2vv_{xx} + 2(v_x)^2 + 2(u_y)^2 + 2uu_{yy} + 2vv_{yy} + 2(v_y)^2 \\ &= 2(u_x)^2 + 2(v_x)^2 + 2(u_y)^2 + 2(v_y)^2 + 2u(u_{xx} + u_{yy}) + 2v(v_{xx} + v_{yy}) \\ &= 2(u_x)^2 + 2(v_x)^2 + 2(u_y)^2 + 2(v_y)^2 \end{align*}\] as \(u\) and \(v\) are harmonic by part 3. This means that \(u_x\), \(u_y\), \(v_x\), and \(v_y\) are all zero, so \(f\) is a constant function.

Conversely it is trivial that \(f\) constant implies \(\left\lvert f\right\rvert^2\) harmonic. Therefore \(\left\lvert f\right\rvert^2\) is harmonic iff \(f\) is constant.

Exercise 3

1. No. Let \(f(x,y) = e^{(x^2 + y^2)} + \cos(x^2 + y^2) - \sin(x + iy) + 2\) Then, \[\frac{\partial f}{\partial \overline{z}} = \frac{1}{2}(\frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y}) = 2(x+iy)(e^{(x^2 + y^2)} - \cos(x^2 + y^2))\] which is non-zero.

2. No. The function is not continuous when \(z^4 = -1\) as the numerator is always non-zero and denominator is.

Exercise 4

For \((x,y)\) in unit disc, \[\begin{align*} u(x,y) &= \frac{x+1}{x^2+y^2+2x+1} \\ \frac{\partial u}{\partial x} &= \frac{1}{x^2+y^2+2x+1} - 2\left(\frac{x+1}{x^2+y^2+2x+1}\right)^2 \\ &= \frac{1}{x^2+y^2+2x+1} - 2u^2 \\ \frac{\partial^2u}{\partial x^2} &= - \frac{2x+2}{(x^2+y^2+2x+1)^2} - 4u\frac{\partial u}{\partial x} \\ &= 8u^3 - \frac{6}{x^2+y^2+2x+1}u \\ \frac{\partial u}{\partial y} &= -2\frac{(x+1)y}{(x^2+y^2+2x+1)^2} \\ \frac{\partial^2u}{\partial y^2} &= -2\frac{(x+1)(x^2+y^2+2x+1)^2 - (x+1)y 2g(2y)}{(x^2+y^2+2x+1)^4} \\ &= -2\frac{x+1}{(x^2+y^2+2x+1)^2} + 8 \frac{(x+1)y^2}{(x^2+y^2+2x+1)^3} \\ &= \left(\frac{8y^2 - 2(x^2+y^2+2x+1)}{(x^2+y^2+2x+1)^2}\right)u \end{align*}\] so we have \[\begin{align*} \Delta u &= 8u^3 + u\left( \frac{8y^2 - 8y^2 - 8(x+1)^2}{(x^2+y^2+2x+1)^2} \right) \\ &= 8u^3 - 8u^3 = 0 \end{align*}\] and therefore \(u\) is harmonic.

By Exercise 1.3, for \(v\) is a harmonic conjugate of \(u\), iff \(v\) satisfies \[ \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y} = \frac{2(x+1)y}{(y^2 + (x+1)^2)^2} \] and \[ \frac{\partial v}{\partial y} = \frac{\partial u}{\partial x} = \frac{y^2 - (x+1)^2}{(y^2+(x+1)^2)^2} \] we can verify that a solution of this differential equation is \[ v(x,y) = \frac{-y}{y^2 + (x+1)^2}. \]

Exercise 5

Let \(g(z) = zf(z)\) and \(g(0) = 0\). Then \(g\) is bounded near 0, and thus a removable singularity at 0. But \(\lim_{z \to 0} g(z) = \lim_{z \to 0} |\sqrt{z}| (|\sqrt{z}| f(z)) = 0\). Then, \(g(z) = zh(z)\) for some analytic \(h\). But \(f = h\) on the punctured unit disk, which implies \(f\) is bounded near 0, which is a contradiction.

Exercise 6

By Liouville’s theorem as \(f(z) + e^z - 3z^5 + 5\) is still entire and is also bounded it is constant, therefore \[\begin{align*} f(2) + e^2 - 3(2)^5 + 5 &= f(0) + e^0 + 5 \\ &= 7 \end{align*}\] so \[ f(2) = 29 - e^2. \]

Exercise 7

No. Consider \(g(z) = e^{-z}(1+ f(z))(z+3)\). As \(g\) is holomorphic and \(|g(z)| \leq 1\) when \(z \neq -3\), thus \(g\) is constant by Liouville’s theorem. Then, \(f(z) = ce^z(z+3)^{-1} - 1\) for some \(c\), but \(f\) is continuous everywhere, which is a contradiction as the limit \(\lim_{z \to -3}f(z)\) does not exist unless \(c = 0\).

Exercise 8

Removable singularity at \(z=1\), poles of order 1 at \(z=e^{2k\pi i/3}\) for \(k = 1,2\), and essential singularity at \(z=0\).