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Diffstat (limited to 'source/know/concept/holomorphic-function')
| -rw-r--r-- | source/know/concept/holomorphic-function/index.md | 46 |
1 files changed, 26 insertions, 20 deletions
diff --git a/source/know/concept/holomorphic-function/index.md b/source/know/concept/holomorphic-function/index.md index cf252c0..976758b 100644 --- a/source/know/concept/holomorphic-function/index.md +++ b/source/know/concept/holomorphic-function/index.md @@ -9,13 +9,13 @@ layout: "concept" --- 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. +is called **holomorphic** or **analytic** if it is **complex differentiable** +in the vicinity of every point of its domain. This is a very strong condition. 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. +they are very "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, @@ -23,14 +23,17 @@ and for all directions of $$\Delta z$$: $$\begin{aligned} \boxed{ - f'(z) = \lim_{\Delta z \to 0} \frac{f(z + \Delta z) - f(z)}{\Delta z} + f'(z) + = \lim_{\Delta z \to 0} \frac{f(z + \Delta z) - f(z)}{\Delta z} } \end{aligned}$$ We decompose $$f$$ into the real functions $$u$$ and $$v$$ of real variables $$x$$ and $$y$$: $$\begin{aligned} - f(z) = f(x + i y) = u(x, y) + i v(x, y) + f(z) + = f(x + i y) + = u(x, y) + i v(x, y) \end{aligned}$$ Since we are free to choose the direction of $$\Delta z$$, we choose $$\Delta x$$ and $$\Delta y$$: @@ -56,9 +59,9 @@ $$\begin{aligned} } \end{aligned}$$ -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. +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. @@ -70,7 +73,8 @@ provided that $$f(z)$$ is holomorphic for all $$z$$ in the area enclosed by $$C$ $$\begin{aligned} \boxed{ - \oint_C f(z) \dd{z} = 0 + \oint_C f(z) \dd{z} + = 0 } \end{aligned}$$ @@ -86,34 +90,36 @@ $$\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$$ satisfy the Cauchy-Riemann -equations, such that the integrands disappear and the final result is zero. +Since $$f(z)$$ is holomorphic, $$u$$ and $$v$$ satisfy the Cauchy-Riemann equations, +such that the integrands disappear and the final result is zero. {% include proof/end.html id="proof-int-theorem" %} -An interesting consequence is **Cauchy's integral formula**, which -states that the value of $$f(z)$$ at an arbitrary point $$z_0$$ is -determined by its values on an arbitrary contour $$C$$ around $$z_0$$: +An interesting consequence is **Cauchy's integral formula**, +which states that the value of $$f(z)$$ at an arbitrary point $$z_0$$ +is determined by its values on an arbitrary contour $$C$$ around $$z_0$$: $$\begin{aligned} \boxed{ - f(z_0) = \frac{1}{2 \pi i} \oint_C \frac{f(z)}{z - z_0} \dd{z} + f(z_0) + = \frac{1}{2 \pi i} \oint_C \frac{f(z)}{z - z_0} \dd{z} } \end{aligned}$$ {% include proof/start.html id="proof-int-formula" -%} -Thanks to the integral theorem, we know that the shape and size -of $$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}$$. Then -we integrate by substitution: +Thanks to the integral theorem, we know that +the shape and size of $$C$$ are irrelevant. +Therefore we choose it to be a circle with radius $$r$$, +such that the integration variable becomes $$z = z_0 + r e^{i \theta}$$. +Then we integrate by substitution: $$\begin{aligned} \frac{1}{2 \pi i} \oint_C \frac{f(z)}{z - z_0} \dd{z} |