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Diffstat (limited to 'source/know/concept/holomorphic-function')
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diff --git a/source/know/concept/holomorphic-function/index.md b/source/know/concept/holomorphic-function/index.md index e22799c..5dde240 100644 --- a/source/know/concept/holomorphic-function/index.md +++ b/source/know/concept/holomorphic-function/index.md @@ -8,7 +8,7 @@ categories: layout: "concept" --- -In complex analysis, a complex function $f(z)$ of a complex variable $z$ +In complex analysis, a complex function $$f(z)$$ of a complex variable $$z$$ is called **holomorphic** or **analytic** if it is complex differentiable in the neighbourhood of every point of its domain. This is a very strong condition. @@ -17,9 +17,9 @@ As a result, holomorphic functions are infinitely differentiable and equal their Taylor expansion at every point. In physicists' terms, they are extremely "well-behaved" throughout their domain. -More formally, a given function $f(z)$ is holomorphic in a certain region -if the following limit exists for all $z$ in that region, -and for all directions of $\Delta z$: +More formally, a given function $$f(z)$$ is holomorphic in a certain region +if the following limit exists for all $$z$$ in that region, +and for all directions of $$\Delta z$$: $$\begin{aligned} \boxed{ @@ -27,13 +27,13 @@ $$\begin{aligned} } \end{aligned}$$ -We decompose $f$ into the real functions $u$ and $v$ of real variables $x$ and $y$: +We decompose $$f$$ into the real functions $$u$$ and $$v$$ of real variables $$x$$ and $$y$$: $$\begin{aligned} f(z) = f(x + i y) = u(x, y) + i v(x, y) \end{aligned}$$ -Since we are free to choose the direction of $\Delta z$, we choose $\Delta x$ and $\Delta y$: +Since we are free to choose the direction of $$\Delta z$$, we choose $$\Delta x$$ and $$\Delta y$$: $$\begin{aligned} f'(z) @@ -44,8 +44,8 @@ $$\begin{aligned} = \pdv{v}{y} - i \pdv{u}{y} \end{aligned}$$ -For $f(z)$ to be holomorphic, these two results must be equivalent. -Because $u$ and $v$ are real by definition, +For $$f(z)$$ to be holomorphic, these two results must be equivalent. +Because $$u$$ and $$v$$ are real by definition, we thus arrive at the **Cauchy-Riemann equations**: $$\begin{aligned} @@ -56,7 +56,7 @@ $$\begin{aligned} } \end{aligned}$$ -Therefore, a given function $f(z)$ is holomorphic if and only if its real +Therefore, a given function $$f(z)$$ is holomorphic if and only if its real and imaginary parts satisfy these equations. This gives an idea of how strict the criteria are to qualify as holomorphic. @@ -64,8 +64,8 @@ strict the criteria are to qualify as holomorphic. ## Integration formulas Holomorphic functions satisfy **Cauchy's integral theorem**, which states -that the integral of $f(z)$ over any closed curve $C$ in the complex plane is zero, -provided that $f(z)$ is holomorphic for all $z$ in the area enclosed by $C$: +that the integral of $$f(z)$$ over any closed curve $$C$$ in the complex plane is zero, +provided that $$f(z)$$ is holomorphic for all $$z$$ in the area enclosed by $$C$$: $$\begin{aligned} \boxed{ @@ -78,7 +78,7 @@ $$\begin{aligned} <label for="proof-int-theorem">Proof</label> <div class="hidden" markdown="1"> <label for="proof-int-theorem">Proof.</label> -Just like before, we decompose $f(z)$ into its real and imaginary parts: +Just like before, we decompose $$f(z)$$ into its real and imaginary parts: $$\begin{aligned} \oint_C f(z) \dd{z} @@ -88,21 +88,21 @@ $$\begin{aligned} &= \oint_C u \dd{x} - v \dd{y} + i \oint_C v \dd{x} + u \dd{y} \end{aligned}$$ -Using Green's theorem, we integrate over the area $A$ enclosed by $C$: +Using Green's theorem, we integrate over the area $$A$$ enclosed by $$C$$: $$\begin{aligned} \oint_C f(z) \dd{z} &= - \iint_A \pdv{v}{x} + \pdv{u}{y} \dd{x} \dd{y} + i \iint_A \pdv{u}{x} - \pdv{v}{y} \dd{x} \dd{y} \end{aligned}$$ -Since $f(z)$ is holomorphic, $u$ and $v$ satisfy the Cauchy-Riemann +Since $$f(z)$$ is holomorphic, $$u$$ and $$v$$ satisfy the Cauchy-Riemann equations, such that the integrands disappear and the final result is zero. </div> </div> An interesting consequence is **Cauchy's integral formula**, which -states that the value of $f(z)$ at an arbitrary point $z_0$ is -determined by its values on an arbitrary contour $C$ around $z_0$: +states that the value of $$f(z)$$ at an arbitrary point $$z_0$$ is +determined by its values on an arbitrary contour $$C$$ around $$z_0$$: $$\begin{aligned} \boxed{ @@ -116,8 +116,8 @@ $$\begin{aligned} <div class="hidden" markdown="1"> <label for="proof-int-formula">Proof.</label> Thanks to the integral theorem, we know that the shape and size -of $C$ is irrelevant. Therefore we choose it to be a circle with radius $r$, -such that the integration variable becomes $z = z_0 + r e^{i \theta}$. Then +of $$C$$ is irrelevant. Therefore we choose it to be a circle with radius $$r$$, +such that the integration variable becomes $$z = z_0 + r e^{i \theta}$$. Then we integrate by substitution: $$\begin{aligned} @@ -126,13 +126,14 @@ $$\begin{aligned} = \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + r e^{i \theta}) \dd{\theta} \end{aligned}$$ -We may choose an arbitrarily small radius $r$, such that the contour approaches $z_0$: +We may choose an arbitrarily small radius $$r$$, such that the contour approaches $$z_0$$: $$\begin{aligned} \lim_{r \to 0}\:\: \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + r e^{i \theta}) \dd{\theta} &= \frac{f(z_0)}{2 \pi} \int_0^{2 \pi} \dd{\theta} = f(z_0) \end{aligned}$$ + </div> </div> @@ -153,7 +154,7 @@ $$\begin{aligned} <label for="proof-diff-formula">Proof</label> <div class="hidden" markdown="1"> <label for="proof-diff-formula">Proof.</label> -By definition, the first derivative $f'(z)$ of a +By definition, the first derivative $$f'(z)$$ of a holomorphic function exists and is: $$\begin{aligned} @@ -183,8 +184,8 @@ $$\begin{aligned} = \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{(\zeta - z_0)^2} \dd{\zeta} \end{aligned}$$ -Since the second-order derivative $f''(z)$ is simply the derivative of $f'(z)$, -this proof works inductively for all higher orders $n$. +Since the second-order derivative $$f''(z)$$ is simply the derivative of $$f'(z)$$, +this proof works inductively for all higher orders $$n$$. </div> </div> |