diff options
Diffstat (limited to 'content/know/concept/holomorphic-function/index.pdc')
| -rw-r--r-- | content/know/concept/holomorphic-function/index.pdc | 71 |
1 files changed, 45 insertions, 26 deletions
diff --git a/content/know/concept/holomorphic-function/index.pdc b/content/know/concept/holomorphic-function/index.pdc index 1077060..1c2f092 100644 --- a/content/know/concept/holomorphic-function/index.pdc +++ b/content/know/concept/holomorphic-function/index.pdc @@ -77,8 +77,12 @@ $$\begin{aligned} } \end{aligned}$$ -*__Proof__*. -*Just like before, we decompose $f(z)$ into its real and imaginary parts:* +<div class="accordion"> +<input type="checkbox" id="proof-int-theorem"/> +<label for="proof-int-theorem">Proof</label> +<div class="hidden"> +<label for="proof-int-theorem">Proof.</label> +Just like before, we decompose $f(z)$ into its real and imaginary parts: $$\begin{aligned} \oint_C f(z) \:dz @@ -88,16 +92,17 @@ $$\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) \:dz &= - \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}$$ -*Since $f(z)$ is holomorphic, $u$ and $v$ satisfy the Cauchy-Riemann -equations, such that the integrands disappear and the final result is zero.* -*__Q.E.D.__* +Since $f(z)$ is holomorphic, $u$ and $v$ satisfy the Cauchy-Riemann +equations, such that the integrands disappear and the final result is zero. +</div> +</div> An interesting consequence is **Cauchy's integral formula**, which states that the value of $f(z)$ at an arbitrary point $z_0$ is @@ -109,11 +114,15 @@ $$\begin{aligned} } \end{aligned}$$ -*__Proof__*. -*Thanks to the integral theorem, we know that the shape and size +<div class="accordion"> +<input type="checkbox" id="proof-int-formula"/> +<label for="proof-int-formula">Proof</label> +<div class="hidden"> +<label for="proof-int-formula">Proof.</label> +Thanks to the integral theorem, we know that the shape and size of $C$ is irrelevant. Therefore we choose it to be a circle with radius $r$, such that the integration variable becomes $z = z_0 + r e^{i \theta}$. Then -we integrate by substitution:* +we integrate by substitution: $$\begin{aligned} \frac{1}{2 \pi i} \oint_C \frac{f(z)}{z - z_0} \dd{z} @@ -121,15 +130,15 @@ $$\begin{aligned} = \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + r e^{i \theta}) \dd{\theta} \end{aligned}$$ -*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 $z_0$: $$\begin{aligned} \lim_{r \to 0}\:\: \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + r e^{i \theta}) \dd{\theta} &= \frac{f(z_0)}{2 \pi} \int_0^{2 \pi} \dd{\theta} = f(z_0) \end{aligned}$$ - -*__Q.E.D.__* +</div> +</div> Similarly, **Cauchy's differentiation formula**, or **Cauchy's integral formula for derivatives** @@ -143,16 +152,20 @@ $$\begin{aligned} } \end{aligned}$$ -*__Proof__*. -*By definition, the first derivative $f'(z)$ of a -holomorphic function $f(z)$ exists and is given by:* +<div class="accordion"> +<input type="checkbox" id="proof-diff-formula"/> +<label for="proof-diff-formula">Proof</label> +<div class="hidden"> +<label for="proof-diff-formula">Proof.</label> +By definition, the first derivative $f'(z)$ of a +holomorphic function exists and is: $$\begin{aligned} f'(z_0) = \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0} \end{aligned}$$ -*We evaluate the numerator using Cauchy's integral theorem as follows:* +We evaluate the numerator using Cauchy's integral theorem as follows: $$\begin{aligned} f'(z_0) @@ -166,7 +179,7 @@ $$\begin{aligned} \oint_C \frac{f(\zeta) (z - z_0)}{(\zeta - z)(\zeta - z_0)} \dd{\zeta} \end{aligned}$$ -*This contour integral converges uniformly, so we may apply the limit on the inside:* +This contour integral converges uniformly, so we may apply the limit on the inside: $$\begin{aligned} f'(z_0) @@ -174,9 +187,10 @@ $$\begin{aligned} = \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{(\zeta - z_0)^2} \dd{\zeta} \end{aligned}$$ -*Since the second-order derivative $f''(z)$ is simply the derivative of $f'(z)$, -this proof works inductively for all higher orders $n$.* -*__Q.E.D.__* +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> ## Residue theorem @@ -205,24 +219,29 @@ $$\begin{aligned} } \end{aligned}$$ -*__Proof__*. *From the definition of a meromorphic function, +<div class="accordion"> +<input type="checkbox" id="proof-res-theorem"/> +<label for="proof-res-theorem">Proof</label> +<div class="hidden"> +<label for="proof-res-theorem">Proof.</label> +From the definition of a meromorphic function, we know that we can decompose $f(z)$ like so, -where $h(z)$ is holomorphic and $p$ are all its poles:* +where $h(z)$ is holomorphic and $p$ are all its poles: $$\begin{aligned} f(z) = h(z) + \sum_{p} \frac{R_p}{z - z_p} \end{aligned}$$ -*We integrate this over a contour $C$ which contains all poles, and apply -both Cauchy's integral theorem and Cauchy's integral formula to get:* +We integrate this over a contour $C$ which contains all poles, and apply +both Cauchy's integral theorem and Cauchy's integral formula to get: $$\begin{aligned} \oint_C f(z) \dd{z} &= \oint_C h(z) \dd{z} + \sum_{p} R_p \oint_C \frac{1}{z - z_p} \dd{z} = \sum_{p} R_p \: 2 \pi i \end{aligned}$$ - -*__Q.E.D.__* +</div> +</div> This theorem might not seem very useful, but in fact, thanks to some clever mathematical magic, |