Exercises

Write $f = u + iv$, then $0 = 2 f' = u_x + iv_x = u_y - iu_y$, so $\operatorname{grad}u = \operatorname{grad}v = 0$. Show $f$ is constant along every straight line segment $L$ by computing the directional derivative $\operatorname{grad}u \cdot \mathbf{v} = 0$ along $L$ connecting $p, q$. Then $u(p) = u(q) = a$ some constant, and $v(p) = v(q) = b$, so $f(z) = a+bi$ everywhere.

• Strategy: show $f'=0$.

• Write $f = u + iv$. Since $f$ is analytic, it satisfies CR, so $$\begin{align*} u_x = v_y && u_y = -v_x .\end{align*}$$

• Similarly write $\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu = U + iV$ where $U = u$ and $V = -v$. Since $\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu$ is analytic, it also satisfies CR , so $$\begin{align*} U_x = V_y && U_y = -V_x \\ \\ \implies u_x = -v_y && u_y = v_x .\end{align*}$$

• Add the LHS of these two equations to get $2u_x = 0 \implies u_x = 0$. Subtract the right-hand side to get $-2v_x = 0 \implies v_x = 0$

• Since $f$ is analytic, it is holomorphic, so $f'$ exists and satisfies $f' = u_x + iv_x$. But by above, this is zero.

• By the previous exercise, $f'=0 \implies f$ is constant.

Part 3:

• Write ${\left\lvert {f} \right\rvert} = c \in {\mathbb{R}}$.
• If $c=0$, done, so suppose $c>0$.
• Use $f\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu = {\left\lvert {f} \right\rvert}^2 = c^2$ to write $\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu=c^2/f$.
• Since ${\left\lvert {f(z)} \right\rvert} = 0 \iff f(z) = 0$, we have $f\neq 0$ on $\Omega$, so $\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu$ is analytic.
• Similarly $f$ is analytic, and $f,\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu$ analytic implies $f'=0$ implies $f$ is constant.

$\implies$: Suppose the first $m-1$ derivatives vanish. Then $$\begin{align*} f(z) = \sum_{k\geq 0} c_k (z-a)^k = \sum_{k\geq m+1} c_k (z-a)^k = (z-a)^m \sum_{k\geq m+1} c_k (z-a)^{k-m} = (z-a)^m (c_m + c_{m+1}(z-a) + \cdots) \coloneqq(z-a)^m g(z) ,\end{align*}$$ using that $c_k \approx f^{(k)}(a)$. Noting that $g(a) = c_m \neq 0$, we have $f(z) = (z-a)^m g(z)$ where $g$ is nonvanishing in a neighborhood of $a$, making $a$ a zero of $f$ of order $m$.

Conversely, if $a$ is an order $m$ zero, then $f(z) = (z-a)^m h(z)$ for $h$ nonvanishing near $a$. So as above, writing $$\begin{align*} f(z) = \sum_{k\geq 0} c_k (z-a)^k = \sum_{k\leq m} c_k (z-a)^k + (z-a)^m g(z) ,\end{align*}$$ we have $$\begin{align*} 0 = f(z) - (z-a)^m h(z) = \sum_{k \leq m} c_k (z-a)^k + (z-a)^m(g(z) - h(z)) .\end{align*}$$ But this is a power series expansion of the zero function, and by uniqueness of power series we have $c_k = 0$ for $k\leq m-1$ and $g(z) = h(z)$. In particular, $g(a) = c_{m}$ by definition, and $g(a) = h(a) \neq 0$.

By contradiction: without loss of generality suppose $f(0) = 0$ and $f'$ vanishes at zero. Then $f(z) = \sum_{k\geq 0} c_k z^k = \sum_{k\geq 2}c_k z^k$ since $c_k \approx f^{(k)}(0)$, so $z=0$ is a zero of order at least 2. But then $f(z) = c_2 z^2 + \cdots$, so $f$ is at best 2-to-1 near 0, contradicting injectivity.