1.3: Derivatives
- Page ID
- 74224
Given a function \(f = u+iv\), the complex conjugate is \(\bar{f} = u-iv\), defined simply by \(z \mapsto \overline{f(z)}\). When \(f\) is holomorphic, then \(\bar{f}\) is called an antiholomorphic function. An antiholomorphic function is a function that depends on \(\bar{z}\) but not on \(z\). So if we write the variable, we write \(\bar{f}\) as \(\bar{f}(\bar{z})\). Let us see why this makes sense. Using the definitions of the Wirtinger operators, \[\frac{\partial \bar{f}}{\partial z_j} = \overline{\frac{\partial f}{\partial \bar{z}_j}} = 0, \qquad \frac{\partial \bar{f}}{\partial \bar{z}_j} = \overline{ \left(\frac{\partial f}{\partial z_j} \right) } , \qquad \text{for all $j=1,\ldots,n$.}\]
For functions that are neither holomorphic or antiholomorphic, we pretend they depend on both \(z\) and \(\bar{z}\). Since we want to write functions in terms of \(z\) and \(\bar{z}\), let us figure out how the chain rule works for Wirtinger derivatives, rather than writing derivatives in terms of \(x\) and \(y\).
Complex Chain Rule
Suppose \(U \subset \mathbb{C}^n\) and \(V \subset \mathbb{C}^m\) are open sets and suppose \(f \colon U \to V\), and \(g \colon V \to \mathbb{C}\) are (real) differentiable functions (mappings). Write the variables as \(z = (z_1,\ldots,z_n) \in U \subset \mathbb{C}^n\) and \(w = (w_1,\ldots,w_m) \in V \subset \mathbb{C}^m\). Then for \(j=1,\ldots,n\), \[\begin{align} & \frac{\partial}{\partial z_j} \left[ g \circ f \right] = \sum_{\ell=1}^m \left( \frac{\partial g}{\partial w_\ell} \frac{\partial f_\ell}{\partial z_j} + \frac{\partial g}{\partial \bar{w}_\ell} \frac{\partial \bar{f}_\ell}{\partial z_j} \right),\label{eq:1} \\ & \frac{\partial}{\partial \bar{z}_j} \left[ g \circ f \right] = \sum_{\ell=1}^m \left( \frac{\partial g}{\partial w_\ell} \frac{\partial f_\ell}{\partial \bar{z}_j} + \frac{\partial g}{\partial \bar{w}_\ell} \frac{\partial \bar{f}_\ell}{\partial \bar{z}_j} \right) . \nonumber \end{align}\]
- Proof
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Write \(f = u+iv\), \(z = x+iy\), \(w=s+it\), \(f\) is a function of \(z\), and \(g\) is a function of \(w\). The composition plugs in \(f\) for \(w\), and so it plugs in \(u\) for \(s\), and \(v\) for \(t\). Using the standard chain rule, \[\begin{align}\begin{aligned} \frac{\partial}{\partial z_j} \left[ g \circ f \right] & = \frac{1}{2} \left( \frac{\partial}{\partial x_j} - i \frac{\partial}{\partial y_j} \right) \left[ g \circ f \right] \\ & = \frac{1}{2} \sum_{\ell=1}^m \left( \frac{\partial g}{\partial s_\ell} \frac{\partial u_\ell}{\partial x_j} + \frac{\partial g}{\partial t_\ell} \frac{\partial v_\ell}{\partial x_j} - i \left( \frac{\partial g}{\partial s_\ell} \frac{\partial u_\ell}{\partial y_j} + \frac{\partial g}{\partial t_\ell} \frac{\partial v_\ell}{\partial y_j} \right) \right) \\ & = \sum_{\ell=1}^m \left( \frac{\partial g}{\partial s_\ell} \, \frac{1}{2} \left( \frac{\partial u_\ell}{\partial x_j} - i \frac{\partial u_\ell}{\partial y_j} \right) + \frac{\partial g}{\partial t_\ell} \, \frac{1}{2} \left( \frac{\partial v_\ell}{\partial x_j} - i \frac{\partial v_\ell}{\partial y_j} \right) \right) \\ & = \sum_{\ell=1}^m \left( \frac{\partial g}{\partial s_\ell} \frac{\partial u_\ell}{\partial z_j} + \frac{\partial g}{\partial t_\ell} \frac{\partial v_\ell}{\partial z_j} \right) . \end{aligned}\end{align}\] For \(\ell = 1, \ldots, m\), \[\frac{\partial}{\partial s_\ell} = \frac{\partial}{\partial w_\ell} + \frac{\partial}{\partial \bar{w}_\ell} , \qquad \frac{\partial}{\partial t_\ell} = i \left( \frac{\partial}{\partial w_\ell} - \frac{\partial}{\partial \bar{w}_\ell} \right) .\] Continuing: \[\begin{align}\begin{aligned} \frac{\partial}{\partial z_j} \left[ g \circ f \right] & = \sum_{\ell=1}^m \left( \frac{\partial g}{\partial s_\ell} \frac{\partial u_\ell}{\partial z_j} + \frac{\partial g}{\partial t_\ell} \frac{\partial v_\ell}{\partial z_j} \right) \\ & = \sum_{\ell=1}^m \left( \left( \frac{\partial g}{\partial w_\ell} \frac{\partial u_\ell}{\partial z_j} + \frac{\partial g}{\partial \bar{w}_\ell} \frac{\partial u_\ell}{\partial z_j} \right) + i \left( \frac{\partial g}{\partial w_\ell} \frac{\partial v_\ell}{\partial z_j} - \frac{\partial g}{\partial \bar{w}_\ell} \frac{\partial v_\ell}{\partial z_j} \right) \right) \\ & = \sum_{\ell=1}^m \left( \frac{\partial g}{\partial w_\ell} \left( \frac{\partial u_\ell}{\partial z_j} + i \frac{\partial v_\ell}{\partial z_j} \right) + \frac{\partial g}{\partial \bar{w}_\ell} \left( \frac{\partial u_\ell}{\partial z_j} -i \frac{\partial v_\ell}{\partial z_j} \right) \right) \\ & = \sum_{\ell=1}^m \left( \frac{\partial g}{\partial w_\ell} \frac{\partial f_\ell}{\partial z_j} + \frac{\partial g}{\partial \bar{w}_\ell} \frac{\partial \bar{f}_\ell}{\partial z_j} \right) . \end{aligned}\end{align}\]
The \(\bar{z}\) derivative works similarly.
Because of the proposition, when we deal with a possibly nonholomorphic function \(f\), we often write \(f(z,\bar{z})\) and treat \(f\) as a function of \(z\) and \(\bar{z}\).
It is good to notice the subtlety of what we just said. Formally it seems as if we are treating \(z\) and \(\bar{z}\) as independent variables when taking derivatives, but in reality they are not independent if we actually wish to evaluate the function. Under the hood, a smooth function that is not necessarily holomorphic is really a function of the real variables \(x\) and \(y\), where \(z = x+iy\).
Another remark is that we could have swapped \(z\) and \(\bar{z}\), by flipping the bars everywhere. There is no difference between the two, they are twins in effect. We just need to know which one is which. After all, it all starts with taking the two square roots of \(-1\) and deciding which one is \(i\) (remember the chickens?). There is no “natural choice” for that, but once we make that choice we must be consistent. And once we picked which root is \(i\), we also picked what is holomorphic and what is antiholomorphic. This is a subtle philosophical as much as a mathematical point.
Let \(U \subset \mathbb{C}^n\) be open. A mapping \(f \colon U \to \mathbb{C}^m\) is said to be holomorphic if each component is holomorphic. That is, if \(f = (f_1,\ldots,f_m)\), then each \(f_j\) is a holomorphic function.
As in one variable, the composition of holomorphic functions (mappings) is holomorphic.
Let \(U \subset \mathbb{C}^n\) and \(V \subset \mathbb{C}^m\) be open sets and suppose \(f \colon U \to V\) and \(g \colon V \to \mathbb{C}^k\) are both holomorphic. Then the composition \(g \circ f\) is holomorphic.
- Proof
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The proof is almost trivial by chain rule. Again let \(g\) be a function of \(w \in V\) and \(f\) be a function of \(z \in U\). For any \(j = 1,\ldots,n\) and any \(p=1,\ldots,k\), we compute \[\frac{\partial}{\partial \bar{z}_j} \left[ g_p \circ f \right] = \sum_{\ell=1}^m \left( \frac{\partial g_p}{\partial w_\ell} \cancelto{0}{\frac{\partial f_\ell}{\partial \bar{z}_j}} + \cancelto{0}{\frac{\partial g_p}{\partial \bar{w}_\ell}} \frac{\partial \bar{f}_\ell}{\partial \bar{z}_j} \right) = 0 . \]
For holomorphic functions the chain rule simplifies, and it formally looks like the familiar vector calculus rule. Suppose again \(U \subset \mathbb{C}^n\) and \(V \subset \mathbb{C}^m\) are open, and \(f \colon U \to V\) and \(g \colon V \to \mathbb{C}\) are holomorphic. Name the variables \(z = (z_1,\ldots,z_n) \in U \subset \mathbb{C}^n\) and \(w = (w_1,\ldots,w_m) \in V \subset \mathbb{C}^m\). In formula \(\eqref{eq:1}\) for the \(z_j\) derivative, the \(\bar{w}_j\) derivative of \(g\) is zero and the \(z_j\) derivative of \(\bar{f}_\ell\) is also zero because \(f\) and \(g\) are holomorphic. Therefore, for any \(j=1,\ldots,n\), \[\frac{\partial}{\partial z_j} \left[ g \circ f \right] = \sum_{\ell=1}^m \frac{\partial g}{\partial w_\ell} \frac{\partial f_\ell}{\partial z_j} .\]
Prove using only the Wirtinger derivatives that a holomorphic function that is real-valued must be constant.
Let \(f\) be a holomorphic function on \(\mathbb{C}^n\). When we write \(\bar{f}\) we mean the function \(z \mapsto \overline{f(z)}\), and we usually write \(\bar{f}(\bar{z})\) as the function is antiholomorphic. However, if we write \(\bar{f}(z)\) we really mean \(z \mapsto \overline{f(\bar{z})}\), that is, composing both the function and the argument with conjugation. Prove \(z \mapsto \bar{f}(z)\) is holomorphic, and prove \(f\) is real-valued on \(\mathbb{R}^n\) (when \(y=0\)) if and only if \(f(z) = \bar{f}(z)\) for all \(z \in \mathbb{C}\).
For a \(U \subset \mathbb{C}^n\), a holomorphic mapping \(f \colon U \to \mathbb{C}^m\), and a point \(p \in U\), define the holomorphic derivative, sometimes called the Jacobian matrix, \[Df(p) \overset{\text{def}}{=} \left[ \frac{\partial f_j}{\partial z_k} (p) \right]_{jk} .\] The notation \(f'(p) = Df(p)\) is also used.
Suppose \(U \subset \mathbb{C}^n\) is open, \(\mathbb{R}^n\) is naturally embedded in \(\mathbb{C}^n\). Consider a holomorphic mapping \(f \colon U \to \mathbb{C}^m\) and suppose that \(f|_{U \cap \mathbb{R}^n}\) maps into \(\mathbb{R}^m \subset \mathbb{C}^m\). Prove that given \(p \in U \cap \mathbb{R}^n\), the real Jacobian matrix at \(p\) of the map \(f|_{U \cap \mathbb{R}^n} \colon U \cap \mathbb{R}^n \to \mathbb{R}^m\) is equal to the holomorphic Jacobian matrix of the map \(f\) at \(p\). In particular, \(Df(p)\) is a matrix with real entries.
By the holomorphic chain rule above, as in the theory of real functions, the derivative of the composition is the composition of derivatives (multiplied as matrices).
Chain Rule for Holomorphic Mappings
Let \(U \subset \mathbb{C}^n\) and \(V \subset \mathbb{C}^m\) be open sets. Suppose \(f \colon U \to V\) and \(g \colon V \to \mathbb{C}^k\) are both holomorphic, and \(p \in U\). Then \[D(g \circ f)(p) = Dg\bigl(f(p)\bigr) \, Df(p) .\]
In shorthand, we often simply write \(D(g \circ f) = Dg Df\).
Prove the proposition.
Suppose \(U \subset \mathbb{C}^n\), \(p \in U\), and \(f \colon U \to \mathbb{C}^m\) is a differentiable function at \(p\). Since \(\mathbb{C}^n\) is identified with \(\mathbb{R}^{2n}\), the mapping \(f\) takes \(U \subset \mathbb{R}^{2n}\) to \(\mathbb{R}^{2m}\). The normal vector calculus Jacobian at \(p\) of this mapping (a \(2m \times 2n\) real matrix) is called the real Jacobian, and we write it as \(D_{\mathbb{R}} f (p)\).
Let \(U \subset \mathbb{C}^n\) be an open set, \(p \in U\), and \(f \colon U \to \mathbb{C}^n\) be holomorphic. Then \[|\text{det }Df(p)|^{2}=\text{det}D_{\mathbb{R}}f(p).\]
The expression \(\det D f(p)\) is called the Jacobian determinant and clearly it is important to know if we are talking about the holomorphic Jacobian determinant or the standard real Jacobian determinant \(\det D_{\mathbb{R}} f(p)\). Recall from vector calculus that if the real Jacobian determinant \(\det D_{\mathbb{R}} f(p)\) of a smooth mapping is positive, then the mapping preserves orientation. In particular, the proposition says that holomorphic mappings preserve orientation.
- Proof
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The real mapping using our identification is \((\Re f_1,\Im f_1, \ldots, \Re f_n, \Im f_n)\) as a function of \((x_1,y_1,\ldots,x_n,y_n)\). The statement is about the two Jacobians at \(p\), that is, the derivatives at \(p\). Hence, we can assume that \(p=0\) and \(f\) is complex linear, \(f(z) = Az\) for some \(n \times n\) matrix \(A\). It is just a statement about matrices. The matrix \(A\) is the (holomorphic) Jacobian matrix of \(f\). Let \(B\) be the real Jacobian matrix of \(f\).
Let us change the basis of \(B\) to be \((z,\bar{z})\) using \(z = x+iy\) and \(\bar{z}=x-iy\) on both the target and the source. The change of basis is some invertible complex matrix \(M\) such that \(M^{-1} B M\) (the real Jacobian matrix \(B\) in these new coordinates) is a matrix of the derivatives of \((f_1,\ldots,f_n,\bar{f}_1,\ldots,\bar{f}_n)\) in terms of \((z_1,\ldots,z_n,\bar{z}_1,\ldots,\bar{z}_n)\). In other words, \[M^{-1} B M = \begin{bmatrix} A & 0 \\ 0 & \overline{A} \end{bmatrix} .\] Thus \[\begin{align}\begin{aligned} \det (B) = \det(M^{-1}M B) = \det(M^{-1} B M) \\ = \det(A) \det(\overline{A}) = \det(A) \, \overline{\det(A)} = |\det(A)|^2 . \end{aligned}\end{align}\]
The regular implicit function theorem and the chain rule give that the implicit function theorem holds in the holomorphic setting. The main thing to check is to check that the solution given by the standard implicit function theorem is holomorphic, which follows by the chain rule.
Implicit Function Theorem
Let \(U \subset \mathbb{C}^{n} \times \mathbb{C}^{m}\) be an open set, let \((z,w) \in \mathbb{C}^n \times \mathbb{C}^m\) be our coordinates, and let \(f \colon U \to \mathbb{C}^m\) be a holomorphic mapping. Let \((z^0,w^0) \in U\) be a point such that \(f(z^0,w^0) = 0\) and such that the \(m \times m\) matrix \[\left[ \frac{\partial f_j}{\partial w_k} (z^0,w^0) \right]_{jk}\] is invertible. Then there exists an open set \(V \subset \mathbb{C}^n\) with \(z^0 \in V\), open set \(W \subset \mathbb{C}^m\) with \(w^0 \in W\), \(V \times W \subset U\), and a holomorphic mapping \(g \colon V \to W\), with \(g(z^0) = w^0\) such that for every \(z \in V\), the point \(g(z)\) is the unique point in \(W\) such that \[f\bigl(z,g(z)\bigr) = 0 .\]
Prove the holomorphic implicit function theorem above. Hint: Check that the normal implicit function theorem for \(C^1\) functions applies, and then show that the \(g\) you obtain is holomorphic.
State and prove a holomorphic version of the inverse function theorem.