3. Week 3: Complex Derivatives
≪ 2. Week 2: More on \(\exp\) and \(\log\) | Table of Contents | 4. Week 4: Conformal Mappings ≫The main issue of figuring out what it means to be “complex differentiable” is that there are three mathematically distinct ways of understanding what it means for a function \(f: \mathbb{C}\to \mathbb{C}\) to be differentiable.
Writing \(f(x+iy) = u(x+iy) + i v(x+iy)\), where \(u:\mathbb{C}\to \mathbb{R}\) is the real part of \(f\) and \(v:\mathbb{C}\to \mathbb{R}\) is the imaginary part, we have the following notions of differentiability:
- Partial differentiability: we can understand the partial derivatives \(\frac{\partial u }{\partial x}\), \(\frac{\partial u }{\partial y}\), \(\frac{\partial v}{\partial x}\), and \(\frac{\partial v }{\partial y}\).
- Complex differentiability: we can make sense of the difference quotient \[\lim _{h\to 0} \frac{f(z+h) - f(z)}{h}.\]
- Fréchet differentiability, which interprets the derivative of \(f\) at a point as a \(\mathbb{R}\)-linear approximation to \(f\) near a point.
You do not need to know exactly how the three above definitions relate to each other, but you can imagine that the three ideas are very different. It turns out that it’s not hard to make them all work together, and that’s what we call the Cauchy-Riemann equations.
Definition 1.
A function \(f:\mathbb{C}\to \mathbb{C}\) is complex differentiable (or analytic, or holomorphic, more or less) at a point \(z\in \mathbb{C}\) if it has continuous partial derivatives at all points sufficiently close to \(z\) and satifsies the Cauchy-Riemann equations \[\begin{align*} \frac{\partial u }{\partial x} = \frac{\partial v}{\partial y} && \textrm{and} && \frac{\partial u }{\partial v} = - \frac{\partial v}{\partial x}\end{align*}\] at \(z\). (Here, \(u\) and \(v\) are the real and imaginary parts of \(f\), as before.)
In fact, any function satisfying the above will automatically be “differentiable” as defined in statements 2 and 3 above!
Thus, to verify that a function is complex differentiable (or show that it isn’t), it’s enough to compute the four partial derivatives and determine whether or not the Cauchy-Riemann equations are satisfied. Conversely, if you know that a function is holomorphic, then you can use whichever version of differentiability you want (though often, partial derivatives are easiest to work with).
Example 2.
- Show that \(f(z) = \overline{z}\) is not holomorphic anywhere on \(\mathbb{C} \).
- Show that if \(D\) is a domain on \(\mathbb{C}\) and \(f:D\to \mathbb{C}\) is holomorphic, then \(u=\Re f\) and \(v=\Im f\) are both harmonic. That is, \(\frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2} = 0\) on \(D\) and likewise for \(v\).
- Show that if \(D\) is a domain on \(\mathbb{C} \) and \(f:D\to \mathbb{C}\) is holomorphic and is never zero, then \(g : D\to \mathbb{R}\) given by \(g(z) = \log \left\lvert f(z) \right\rvert\) is harmonic.
Solutions
The whole point of this example is to get familiar with the Cauchy-Riemann equations.
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We have \(f(x+iy) = x - iy\), so writing \(u=\Re f\) and \(v = \Im f\) as before, we have that \(\frac{\partial u }{\partial x} = 1\) while \(\frac{\partial v}{\partial y} = -1\). Since \(\frac{\partial u }{\partial x} \neq \frac{\partial v}{ \partial y}\) anywhere, \(f\) cannot be holomorphic.
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I’ll only do the computations for \(u\), the computations for \(v\) are similar. We have that \[\frac{\partial^2 u }{\partial x^2} = \frac{\partial }{\partial x} \frac{\partial u }{\partial x} = \frac{\partial }{\partial x} \frac{\partial v }{\partial y} = \frac{\partial^2 v}{\partial y \partial x}.\] We used the Cauchy-Riemann equations to get the second equality; we used Clairaut’s theorem to swap the derivatives at the end. Likewise, we have \[\frac{\partial^2 u}{\partial y^2} = \frac{\partial }{\partial y} \frac{\partial u }{\partial y} = - \frac{\partial}{\partial y} \frac{\partial v }{\partial x} = - \frac{\partial^2v }{\partial y \partial x}.\] Putting the two together, we get that \[\frac{\partial^2 u }{\partial x^2} + \frac{\partial u^2}{ \partial y^2} = \frac{\partial^2 v}{\partial y \partial x} - \frac{\partial^2 v}{\partial y \partial x} = 0.\] Hence \(u\) is harmonic. (The proof for \(v\) goes the same way.) \(\square\)
The last problem uses a lengthier computation, but can still be proven by using the chain rule to extract the right partial derivatives. I’ll leave it as a good exercise (though I may have done this in class).
Remark 3. Cauchy-Riemann Equations as Conditions on the Jacobian
If \(f:\mathbb{C}\to \mathbb{C}\) is a complex-differentiable function as described above, we may interpret the Cauchy-Riemann equations in terms of the Jacobian of \(f\), which is \[Df = \begin{pmatrix} \partial_xu & \partial_yu \\ \partial_xv & \partial_yv \end{pmatrix}.\] Here, \(u\) and \(v\) are the real and imaginary parts of \(f\), as before. The equations allow us to rewrite the \(v\)-partials, thus expressing \[Df = \begin{pmatrix} \partial_xu & \partial_yu \\ -\partial_y u & \partial_x u \end{pmatrix}.\] This is a composition of a rotation matrix with a scaling matrix. But at the same time, \(f’(z)\), which is a single complex number, can be interpreted as a linear function \(\mathbb{C}\to \mathbb{C}\). This linear function is exactly a composition of a rotation with a scaling.
Thus, the Cauchy-Riemann equations are in a sense identifying geometric constraints on what the Jacobian could possibly be — \(Df\) needs to mimick the action of a complex number!
For the remainder of class, we’ll be discussing the inverse function theorem. You might remember the single-variable case from calculus: the derivative of the inverse of a function is the inverse (i.e., reciprocal) of the function’s derivative. Of course, this no longer works if the function’s derivative was zero. This generalises to the multivariable setting, though we shall only state it for functions in two variables.
Theorem 4. Inverse Function Theorem
Let \(U\) and \(V\) be two open subsets of \(\mathbb{R}^2\). Let \(f:U\to V\) have continuous first-order partial derivatives, and define the Jacobian matrix of \(f\) as \[Df = \begin{pmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{ \partial y} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2 }{\partial y} \end{pmatrix}.\] Here, \(f(x, y) = \left( f_1(x, y), f_2(x, y) \right)\). Suppose \(Df\) is an invertible matrix at every point in \(U\), and suppose \(f\) has an inverse \(g: V\to U\). Then, \(g\) has continuous first-order partial derivatives everywhere on \(V\), and \(Dg(x, y) = \left( Df\left( g(x, y) \right) \right) ^{-1}\).
This is not really the full version of the inverse function theorem, but for the purposes of this discussion, this will suffice. This statement is bulky enough as it is!
The upshot of the theorem is that the heuristic “the derivative of the inverse is the inverse of the derivative” does generalise to higher-dimensional calculus. However, this comes at the price of having to invert a Jacobian matrix rather than a single real number.
Recall from linear algebra that \(\mathbb{C}\cong \mathbb{R}^2\) as real vector spaces. This means that we should be able to rephrase this in terms of complex-differentiable functions. But the above theorem still holds even when \(f:U\to V\) (where \(U, V\) are now open subsets of \(\mathbb{C}\)) is merely has continuous partial derivatives. What changes when \(f\) is assumed to be complex differentiable?
Theorem 5. Holomorphic Inverse Function Theorem
Let \(U, V\) be open subsets of \(\mathbb{C}\). Suppose \(f:U\to V\) is complex differentiable, that \(f’(z)\neq 0\) for any \(z\in U\), and that \(f\) has an inverse \(g:V\to U\). Then \(g\) is also complex differentiable, and \(g’(w) = \frac{1}{f’\left( g(w) \right)}.\)
We’ll omit the proofs here, but assuming the real version of the inverse function theorem, one can prove the holomorphic version by making the linear isomorphism \(\mathbb{C}\cong \mathbb{R}^2\) explicit and tracking down what happens.
Exercise 6.
Using the inverse function theorem, prove that any branch of the complex logarithm \(\operatorname{Log}\) is holomorphic everywhere on its domain, and that it is an antiderivative of \(\frac{1}{z}\).