# Exercises

Show that if $$f' = 0$$ on a domain $$\Omega$$, then $$f$$ is constant on $$\Omega$$

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.

Show that if $$f$$ and $$\overline{f}$$ are both holomorphic on a domain $$\Omega$$, then $$f$$ is constant on $$\Omega$$.

• 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 $$\overline{f} = U + iV$$ where $$U = u$$ and $$V = -v$$. Since $$\overline{f}$$ 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.

If $$f$$ is holomorphic on $$\Omega$$ and any of the following hold, then $$f$$ is constant:

• $$\Re(f)$$ is constant.
• $$\Im(f)$$ is constant.
• $${\left\lvert {f} \right\rvert}$$ is constant.

Part 3:

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

Show that if $$f$$ is holomorphic in $${\mathbb{D}}_r(a)$$ and $$a$$ is a zero of $$f$$ of multiplicity $$n$$, then $$f^{(k)}(a) = 0$$ for $$k\leq n-1$$ and $$f^{(n)}(a) \neq 0$$. Show that this is an iff.

$$\implies$$: 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 $$g(a) = h(a) \neq 0$$.

Show that if $$f$$ is holomorphic on $$\Omega$$ and injective, then $$f'(z)$$ is nonvanishing on $$\Omega$$.

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.

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