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 \(\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu\) are both holomorphic on a domain \(\Omega\), then \(f\) is constant on \(\Omega\).
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Strategy: show \(f'=0\).
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Write \(f = u + iv\). Since \(f\) is analytic, it satisfies CR, so \begin{align*} u_x = v_y && u_y = -v_x .\end{align*}
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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*}
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Add the LHS of these two equations to get \(2u_x = 0 \implies u_x = 0\). Subtract the right-hand side to get \(-2v_x = 0 \implies v_x = 0\)
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Since \(f\) is analytic, it is holomorphic, so \(f'\) exists and satisfies \(f' = u_x + iv_x\). But by above, this is zero.
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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 \({\left\lvert {f} \right\rvert} = c \in {\mathbb{R}}\).
- 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.
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.