Complex Analysis MT23, Complex differentiability
Flashcards
@Define what it means for a function $f : U \to \mathbb C$ to be complex differentiable.
Let $a \in \mathbb C$ and suppose $f : U \to \mathbb C$ is a function where $U$ is a neighbourhood of $a$ (i.e. $f$ is defined on some ball $B(a, \varepsilon)$). Then $f$ is complex differentiable at $a$ if
\[\lim _ {z \to a} \frac{f(z) - f(a)}{z - a}\]exists.
@Define what it means for a function $f : U \to \mathbb C$ to be complex differentiable with derivative $f'(a)$, using the $\varepsilon(z)$ definition.
Let $a \in \mathbb C$, let $U$ be a neighbourhood of $a$ and $f : U \to \mathbb C$ be a function. Then $f$ is differentiable at $a$ if and only if
\[f(z) = f(a) + f'(a)(z-a) + \varepsilon(z)(z-a)\]for some $\varepsilon(z) \to 0$ as $z \to a$.
Let $U \subseteq \mathbb C$ be an open set. @Define what it means for a function $f : U \to \mathbb C$ to be holomorphic.
$f$ is complex differentiable at every $a \in U$.
As a consequence of Cauchy’s theorem, what is true about how any entire function $f$ can be represented as a power series?
for all $z \in \mathbb C$.
@State the theorem that links the derivative of $f : U \to \mathbb C$ to its partial derivatives.
Let $a \in \mathbb C$ and let $U$ be a neighbourhood of $a$, and let $f : U \to \mathbb C$ be a function which is complex differentiable at $a$. Let $u, v : \mathbb R^2 \to \mathbb R$ be the components of $f$. Then the four partial derivatives $\partial _ x u$, $\partial _ y u$, $\partial _ x v$, $\partial _ y v$ exist at $a$, and moreover
\[\partial _ x u = \partial _ y v\]and
\[\partial _ x v = -\partial _ y u\]and
\[f'(a) = \partial _ xu(a) + i \partial _ x v(a)\]@State the Cauchy-Riemann equations in terms of $\partial _ x u$, $\partial _ y u$, $\partial _ x v$, $\partial _ y v$ (the partial derivatives of the components of $f$).
and
\[\partial _ x v = -\partial _ y u\]@Prove that if
- $f : U \to \mathbb C$ is holomorphic at $a$
- $f$ has components $(u, v)$
then:
- All the partial derivatives $\partial _ x u$, $\partial _ y u$, $\partial _ x v$, $\partial _ y v$ exist
- They satisfy the Cauchy-Riemann equations, i.e.
\[\partial _ x u = \partial _ y v\]
and
\[\partial _ x v = -\partial _ y u\]
Use (or for a shorter proof, do not use) the following notation:
- $f = (u, v)$
- $a = (a _ 1, a _ 2)$
- $h = (h _ 1, h _ 2)$
- $f'(a) = (b _ 1, b _ 2)$
- $\varepsilon = (\varepsilon _ 1, \varepsilon _ 2)$
Proof 1: We use the following characterisation of differentiability at $z = a$: we can write
\[f(a + h) = f(a) + f'(a)h + \varepsilon(h)h\]where $\varepsilon(h) \to 0$ as $h \to 0$. Then, writing from the perspective of $\mathbb R^2$,
\[f(a + h) = \begin{pmatrix}u(a _ 1 + h _ 1, a _ 2 + h _ 2) \\ v(a _ 1 + h _ 1, a _ 2 + h _ 2)\end{pmatrix} = \begin{pmatrix}u(a _ 1, a _ 2) \\ v(a _ 1, a _ 2) \end{pmatrix} + \begin{pmatrix} b _ 1 \\ b _ 2 \end{pmatrix} \begin{pmatrix} h _ 1 \\ h _ 2 \end{pmatrix} + \begin{pmatrix} \varepsilon _ 1(h _ 1, h _ 2) \\ \varepsilon _ 2(h _ 1, h _ 2) \end{pmatrix} \begin{pmatrix} h _ 1 \\ h _ 2 \end{pmatrix}\]Multiplication in $\mathbb C$,
\[\begin{pmatrix}u(a _ 1 + h _ 1, a _ 2 + h _ 2) \\ v(a _ 1 + h _ 1, a _ 2 + h _ 2)\end{pmatrix} = \begin{pmatrix}u(a _ 1, a _ 2) \\ v(a _ 1, a _ 2) \end{pmatrix} + \begin{pmatrix} b _ 1 h _ 1 - b _ 2 h _ 2 \\ b _ 1 h _ 2 + b _ 2 h _ 1 \end{pmatrix} + \begin{pmatrix} \varepsilon _ 1(h _ 1, h _ 2) h _ 1 - \varepsilon _ 2(h _ 1, h _ 2)h _ 2 \\ \varepsilon _ 2(h _ 1, h _ 2)h _ 1 + \varepsilon _ 1(h _ 1, h _ 2) h _ 2 \end{pmatrix}\]Then consider components separately and take turns setting $h _ 1$ and $h _ 2$ to $0$. E.g.
\[u(a _ 1 + h _ 1, a _ 2) = u(a _ 1, a _ 2) + b _ 1 h _ 1 + \varepsilon _ 1(h _ 1, 0) h _ 1\]Shows that $\partial _ x u$ exists and equals $b _ 1$.
Proof 2: (Shorter but slightly less rigorous)
We have that
\[f'(z) = \lim _ {h \to 0} \frac{f(z+h) - f(z)}{h}\]is defined for all $z \in U$. Since all the limits must agree no matter the direction $h$ tends to zero from, if $h$ is real and $h \to 0$, we have
\[\begin{aligned} f'(z) &= \lim _ {h \to 0} \frac{f(z+h) - f(z)}{h} \\ &= \lim _ {h \to 0} \frac{u(x + h, y) + iv(x + h, y) - u(x, y) - iv(x, y)}{h} \\ &= \partial _ x u(x, y) + i\partial _ x v(x, y) \end{aligned}\]If instead $h = i\delta$ for some $\delta$ real and $\delta \to 0$, we have
\[\begin{aligned} f'(z) &= \lim _ {h \to 0} \frac{f(z+h) - f(z)}{h} \\ &= \lim _ {\delta \to 0} \frac 1 i \frac{u(x, y+\delta) + iv(x, y + \delta) - u(x, y) - iv(x, y)}{\delta} \\ &= \frac 1 i (\partial _ y u(x, y) + i\partial _ y v(x, y)) \end{aligned}\]Then setting these equal to one another:
\[\partial _ x u(x, y) + i\partial _ x v(x, y) = -i(\partial _ y u(x, y) + i\partial _ y v(x, y))\]Taking real and imaginary parts, we have the CR equations:
\[\partial _ x u = \partial _ y v\]and
\[\partial _ x v = -\partial _ y u\]@State the constancy theorem for complex functions.
Suppose $f : \mathbb C \to \mathbb C$ is holomorphic and that $f'$ is identically zero. Then $f$ is constant.
@Justify why the following is true: if $f : \mathbb C \to \mathbb C$ is holomorphic and that $f'$ is identically zero, then $f$ is constant.
Consider $f$ as $f = u + i v$, then apply the constancy theorem for real functions to $u$ and $v$ respectively.
Suppose $f : U \to \mathbb C$ is a function with components $(u, v)$, and suppose that the partial derivatives exist. @Define the Wirtinger partial derivatives.
and
\[\partial _ \bar z f = \frac{1}{2} (\partial _ x + i \partial _ y) u + i \frac{1}{2} (\partial _ x + i \partial _ y) v\]The Wirtinger derivatives are defined as:
\[\partial _ z f = \frac{1}{2} (\partial _ x - i \partial _ y) u + i \frac{1}{2}(\partial _ x - i\partial _ y)v\]
and
\[\partial _ \bar z f = \frac{1}{2} (\partial _ x + i \partial _ y) u + i \frac{1}{2} (\partial _ x + i \partial _ y) v\]
(supposing that $f : U \to \mathbb C$ is a function with components $(u, v)$, and that the partial derivatives exist). @State the Cauchy-Riemann equations in this context.
$f$ satisfies the Cauchy-Riemann equations if and only if $\partial _ \bar z f = 0$.
@Define the Laplacian of a function $u : \mathbb R^2 \to \mathbb R$ which is twice differentiable on some open set $U \subseteq \mathbb R^2$.
@Define what it means for a function $u : \mathbb R^2 \to \mathbb R$ which is twice differentiable on some open set $U \subseteq \mathbb R^2$ to be harmonic.
@State and then prove the theorem that links a function being holomorphic with its components being harmonic.
Let $U \subseteq \mathbb C$ be open, and suppose that $f : U \to \mathbb C$ is holomorphic. Let $(u, v)$ be the components of $f$. Then $u$ and $v$ are harmonic
This follows from the Cauchy-Riemann equations, since
\[\partial _ {xx} u = \partial _ {xy} v\]and
\[\partial _ {yy} u = -\partial _ {yx} v\]plus the symmetry property of continuous partial derivatives.
@Define what it means for $u, v : \mathbb R^2 \to \mathbb R$ to be harmonic conjugates.
is holomorphic.
Roughly explain the intuition behind why $f$ being complex differentiable is a much stronger property than just being differentiable as a function of $\mathbb R^2$.
For $f : \mathbb R^2 \to \mathbb R^2$, the property of being differentiable at $\pmb x _ 0$ means that it can be well-approximated by a linear map about $\pmb x _ 0$:
\[f(x) \approx f(\pmb x _ 0) + A(\pmb x - \pmb x _ 0) + o(\pmb x - \pmb x _ 0)\]For $f : \mathbb C \to \mathbb C$, the property is
\[f(z) \approx f(z _ 0) + c(z - z _ 0) + o(z - z _ 0)\]where $c \in \mathbb C$. But scalar multiplication in $\mathbb C$ corresponds to rotations and scaling, so the types of linear map is much more restrictive.
@Prove that the function
\[f(z) = \sum^\infty _ {k=0} \frac{1}{(k-z)^2}\]
converges to a holomorphic function on all of $\mathbb C \setminus \mathbb Z$.
The overall approach is as follows:
- Being holomorphic is a local property, so it suffices to show that for any $z \in \mathbb C \setminus \mathbb Z$, there is a neighbourhood around which $f$ is holomorphic.
- Being holomorphic will follow from being a uniform limit of holomorphic functions in that neighbourhood.
- We want to compare to just $\sum 1/z^2$. The difficulty is that $(k - z)^2$ we might be too close to zero which can mess up this comparison. So we split the sum into two sums, one where all the terms are close to integers, and one where we can guarantee the integers are far away enough.
Fix $z \in \mathbb C \setminus \mathbb Z$. Let $n$ be such that $ \vert z \vert < n$. Define
\[U = \{w \in \mathbb C \setminus \mathbb Z \mid \vert w \vert < n\}\]This is the neighbourhood we use. Then
\[f(z) = \sum^{2n}_ {k = 0} \frac{1}{(k-z)^2} + \sum^\infty_ {k = 2n + 1} \frac{1}{(k-z)^2}\]We can focus on just the second term since the first one is holomorphic for all $w \in U$ as it is just a finite sum. In the second, we have the additional bit of information that $k > 2n$ in all the terms, and for all $w \in U$, this means that $ \vert w \vert < n < k/2$. Hence
\[ \vert k -w \vert ^2 \ge \vert \vert k \vert - \vert w \vert \vert ^2 \ge \vert k - k/2 \vert ^2 = \vert k \vert ^2/4\](with the first inequality following from the reverse triangle inequality). Then for all terms in this second sum
\[\frac{1}{(k - z)^2} \le \frac{4}{k^2}\]so by the M-test, it converges uniformly. Then we are done.
Proofs
Let $U$ be an open subset of $\mathbb C$. @Prove that $f : U \to \mathbb C$ satisfies the Cauchy-Riemann equations if and only if $\partial _ \bar z f = 0$.
Todo.