복소함수의 Wirtinger 도함수
📂복소해석 복소함수의 Wirtinger 도함수 빌드업 복소 함수 f : C → C f : \mathbb{C} \to \mathbb{C} f : C → C 복소수 z = x + i y z=x+iy z = x + i y x , y ∈ R x,y \in \mathbb{R} x , y ∈ R f f f u , v : R 2 → R u,v : \mathbb{R}^{2} \to \mathbb{R} u , v : R 2 → R f f f
f ( z ) = f ( x , y ) = u ( x , y ) + i v ( x , y )
f(z) = f(x,y) = u(x,y) + i v(x,y)
f ( z ) = f ( x , y ) = u ( x , y ) + i v ( x , y )
그러면 f f f 전미분 은 다음과 같다.
d f = d u + i d v = ( ∂ u ∂ x d x + ∂ u ∂ y d y ) + i ( ∂ v ∂ x d x + ∂ v ∂ y d y ) = ( ∂ u ∂ x + i ∂ v ∂ x ) d x + ( ∂ u ∂ y + i ∂ v ∂ y ) d y
\begin{equation}
\begin{aligned}
df &= du + i dv
\\ &= \left( \dfrac{\partial u}{\partial x}dx + \dfrac{\partial u}{\partial y}dy \right) + i \left( \dfrac{\partial v}{\partial x}dx + \dfrac{\partial v}{\partial y}dy \right)
\\ &= \left( \dfrac{\partial u}{\partial x} + i\dfrac{\partial v}{\partial x} \right)dx + \left( \dfrac{\partial u}{\partial y} + i\dfrac{\partial v}{\partial y} \right)dy
\end{aligned}
\end{equation}
df = d u + i d v = ( ∂ x ∂ u d x + ∂ y ∂ u d y ) + i ( ∂ x ∂ v d x + ∂ y ∂ v d y ) = ( ∂ x ∂ u + i ∂ x ∂ v ) d x + ( ∂ y ∂ u + i ∂ y ∂ v ) d y
d z dz d z d z ˉ d\bar{z} d z ˉ d x , d y dx, dy d x , d y
{ d z = d x + i d y d z ˉ = d x − i d y ⟹ { d x = d z + d z ˉ 2 d y = d z − d z ˉ 2 i
\begin{cases}
dz = dx + idy
\\ d\bar{z} = dx -idy
\end{cases}
\implies
\begin{cases}
dx = \dfrac{dz + d\bar{z}}{2}
\\ dy = \dfrac{dz - d\bar{z}}{2i}
\end{cases}
{ d z = d x + i d y d z ˉ = d x − i d y ⟹ ⎩ ⎨ ⎧ d x = 2 d z + d z ˉ d y = 2 i d z − d z ˉ
이를 ( 1 ) (1) ( 1 )
d f = ( ∂ u ∂ x + i ∂ v ∂ x ) d x + ( ∂ u ∂ y + i ∂ v ∂ y ) d y = ( ∂ u ∂ x + i ∂ v ∂ x ) ( d z + d z ˉ 2 ) + ( ∂ u ∂ y + i ∂ v ∂ y ) ( d z − d z ˉ 2 i ) = 1 2 [ ( ∂ u ∂ x + i ∂ v ∂ x ) − i ( ∂ u ∂ y + i ∂ v ∂ y ) ] d z + 1 2 [ ( ∂ u ∂ x + i ∂ v ∂ x ) + i ( ∂ u ∂ y + i ∂ v ∂ y ) ] d z ˉ = 1 2 [ ∂ ( u + i v ) ∂ x − i ∂ ( u + i v ) ∂ y ] d z + 1 2 [ ∂ ( u + i v ) ∂ x + i ∂ ( u + i v ) ∂ y ] d z ˉ = 1 2 [ ∂ f ∂ x − i ∂ f ∂ y ] d z + 1 2 [ ∂ f ∂ x + i ∂ f ∂ y ] d z ˉ = 1 2 [ ∂ ∂ x − i ∂ ∂ y ] f d z + 1 2 [ ∂ ∂ x + i ∂ ∂ y ] f d z ˉ
\begin{align*}
df &= \left( \dfrac{\partial u}{\partial x} + i\dfrac{\partial v}{\partial x} \right)dx + \left( \dfrac{\partial u}{\partial y} + i\dfrac{\partial v}{\partial y} \right)dy
\\ &= \left( \dfrac{\partial u}{\partial x} + i\dfrac{\partial v}{\partial x} \right)\left( \dfrac{dz + d\bar{z}}{2} \right) + \left( \dfrac{\partial u}{\partial y} + i\dfrac{\partial v}{\partial y} \right)\left( \dfrac{dz - d\bar{z}}{2i} \right)
\\ &= \dfrac{1}{2} \left[ \left( \dfrac{\partial u}{\partial x} + i \dfrac{\partial v}{\partial x} \right) -i \left( \dfrac{\partial u}{\partial y} + i\dfrac{\partial v}{\partial y} \right)\right]dz
\\ &\quad+ \dfrac{1}{2} \left[ \left( \dfrac{\partial u}{\partial x} + i \dfrac{\partial v}{\partial x} \right) + i \left( \dfrac{\partial u}{\partial y} + i\dfrac{\partial v}{\partial y} \right)\right]d\bar{z}
\\ &= \dfrac{1}{2} \left[ \dfrac{\partial (u+iv)}{\partial x} - i \dfrac{\partial (u+iv)}{\partial y}\right]dz + \dfrac{1}{2} \left[ \dfrac{\partial (u+iv)}{\partial x} + i \dfrac{\partial (u+iv)}{\partial y}\right]d\bar{z}
\\ &= \dfrac{1}{2} \left[ \dfrac{\partial f}{\partial x} - i \dfrac{\partial f}{\partial y}\right]dz + \dfrac{1}{2} \left[ \dfrac{\partial f}{\partial x} + i \dfrac{\partial f}{\partial y}\right]d\bar{z}
\\ &= \dfrac{1}{2} \left[ \dfrac{\partial }{\partial x} - i \dfrac{\partial }{\partial y}\right]fdz + \dfrac{1}{2} \left[ \dfrac{\partial }{\partial x} + i \dfrac{\partial }{\partial y}\right]fd\bar{z}
\end{align*}
df = ( ∂ x ∂ u + i ∂ x ∂ v ) d x + ( ∂ y ∂ u + i ∂ y ∂ v ) d y = ( ∂ x ∂ u + i ∂ x ∂ v ) ( 2 d z + d z ˉ ) + ( ∂ y ∂ u + i ∂ y ∂ v ) ( 2 i d z − d z ˉ ) = 2 1 [ ( ∂ x ∂ u + i ∂ x ∂ v ) − i ( ∂ y ∂ u + i ∂ y ∂ v ) ] d z + 2 1 [ ( ∂ x ∂ u + i ∂ x ∂ v ) + i ( ∂ y ∂ u + i ∂ y ∂ v ) ] d z ˉ = 2 1 [ ∂ x ∂ ( u + i v ) − i ∂ y ∂ ( u + i v ) ] d z + 2 1 [ ∂ x ∂ ( u + i v ) + i ∂ y ∂ ( u + i v ) ] d z ˉ = 2 1 [ ∂ x ∂ f − i ∂ y ∂ f ] d z + 2 1 [ ∂ x ∂ f + i ∂ y ∂ f ] d z ˉ = 2 1 [ ∂ x ∂ − i ∂ y ∂ ] fd z + 2 1 [ ∂ x ∂ + i ∂ y ∂ ] fd z ˉ
이때 첫번째 항의 상수와 괄호를 ∂ ∂ z \dfrac{\partial }{\partial z} ∂ z ∂ ∂ ∂ z ˉ \dfrac{\partial }{\partial \bar{z}} ∂ z ˉ ∂ f f f
d f = ∂ f ∂ z d z + ∂ f ∂ z ˉ d z ˉ
df = \dfrac{\partial f}{\partial z}dz + \dfrac{\partial f}{\partial \bar{z}}d\bar{z}
df = ∂ z ∂ f d z + ∂ z ˉ ∂ f d z ˉ
정의 x , y ∈ R x,y \in \mathbb{R} x , y ∈ R z = x + i y z=x+iy z = x + i y f : C → C f :\mathbb{C} \to \mathbb{C} f : C → C u , y : R 2 → R u,y : \mathbb{R}^{2} \to \mathbb{R} u , y : R 2 → R f ( z ) = f ( x , y ) = u ( x , y ) + i v ( x , y ) f(z)=f(x,y)=u(x,y)+iv(x,y) f ( z ) = f ( x , y ) = u ( x , y ) + i v ( x , y ) ∂ ∂ z \dfrac{\partial }{\partial z} ∂ z ∂ ∂ ∂ z ˉ \dfrac{\partial }{\partial \bar{z}} ∂ z ˉ ∂
∂ ∂ z : = 1 2 ( ∂ ∂ x − i ∂ ∂ y ) ∂ ∂ z ˉ : = 1 2 ( ∂ ∂ x + i ∂ ∂ y )
\begin{align*}
\dfrac{\partial }{\partial z} & := \dfrac{1}{2} \left( \dfrac{\partial }{\partial x} - i \dfrac{\partial }{\partial y}\right)
\\ \dfrac{\partial }{\partial \bar{z}} & := \dfrac{1}{2} \left( \dfrac{\partial }{\partial x} + i \dfrac{\partial }{\partial y}\right)
\end{align*}
∂ z ∂ ∂ z ˉ ∂ := 2 1 ( ∂ x ∂ − i ∂ y ∂ ) := 2 1 ( ∂ x ∂ + i ∂ y ∂ )
이를 (켤레) Wirtinger 미분 연산자 (conjugate) Wirtinger differential operator 라고 하고, ∂ f ∂ z , ∂ f ∂ z ˉ \dfrac{\partial f}{\partial z}, \dfrac{\partial f}{\partial \bar{z}} ∂ z ∂ f , ∂ z ˉ ∂ f (켤레) Wirtinger 도함수 (conjugate) Wirtinger derivative 라고 한다.
설명 Wirtinger 미분 연산자를 z z z z ˉ \bar{z} z ˉ
∂ z ∂ z = 1 2 ( ∂ ∂ x − i ∂ ∂ y ) ( x + i y ) = 1 2 ( ∂ x ∂ x + ∂ y ∂ y ) = 1
\dfrac{\partial z}{\partial z} = \dfrac{1}{2} \left( \dfrac{\partial }{\partial x} - i \dfrac{\partial }{\partial y}\right)(x+iy) = \dfrac{1}{2} \left( \dfrac{\partial x}{\partial x} + \dfrac{\partial y}{\partial y}\right) = 1
∂ z ∂ z = 2 1 ( ∂ x ∂ − i ∂ y ∂ ) ( x + i y ) = 2 1 ( ∂ x ∂ x + ∂ y ∂ y ) = 1
∂ z ˉ ∂ z ˉ = 1 2 ( ∂ ∂ x + i ∂ ∂ y ) ( x − i y ) = 1 2 ( ∂ x ∂ x + ∂ y ∂ y ) = 1
\dfrac{\partial \bar{z}}{\partial \bar{z}} = \dfrac{1}{2} \left( \dfrac{\partial }{\partial x} + i \dfrac{\partial }{\partial y}\right)(x-iy) = \dfrac{1}{2} \left( \dfrac{\partial x}{\partial x} + \dfrac{\partial y}{\partial y}\right) = 1
∂ z ˉ ∂ z ˉ = 2 1 ( ∂ x ∂ + i ∂ y ∂ ) ( x − i y ) = 2 1 ( ∂ x ∂ x + ∂ y ∂ y ) = 1
∂ z ∂ z ˉ = 1 2 ( ∂ ∂ x + i ∂ ∂ y ) ( x + i y ) = 1 2 ( ∂ x ∂ x − ∂ y ∂ y ) = 0
\dfrac{\partial z}{\partial \bar{z}} = \dfrac{1}{2} \left( \dfrac{\partial }{\partial x} + i \dfrac{\partial }{\partial y}\right)(x+iy) = \dfrac{1}{2} \left( \dfrac{\partial x}{\partial x} - \dfrac{\partial y}{\partial y}\right) = 0
∂ z ˉ ∂ z = 2 1 ( ∂ x ∂ + i ∂ y ∂ ) ( x + i y ) = 2 1 ( ∂ x ∂ x − ∂ y ∂ y ) = 0
∂ z ˉ ∂ z = 1 2 ( ∂ ∂ x − i ∂ ∂ y ) ( x − i y ) = 1 2 ( ∂ x ∂ x − ∂ y ∂ y ) = 0
\dfrac{\partial \bar{z}}{\partial z} = \dfrac{1}{2} \left( \dfrac{\partial }{\partial x} - i \dfrac{\partial }{\partial y}\right)(x-iy) = \dfrac{1}{2} \left( \dfrac{\partial x}{\partial x} - \dfrac{\partial y}{\partial y}\right) = 0
∂ z ∂ z ˉ = 2 1 ( ∂ x ∂ − i ∂ y ∂ ) ( x − i y ) = 2 1 ( ∂ x ∂ x − ∂ y ∂ y ) = 0
위의 결과로부터 Wirtinger 미분 연산자는 마치 z z z z ˉ \bar{z} z ˉ f ( z ) = z ˉ f(z)=\bar{z} f ( z ) = z ˉ f f f z z z d f d z \dfrac{df}{dz} d z df z ˉ \bar{z} z ˉ 정칙함수에 대해서 살펴보면 그 의미가 기존의 미분의 의미와 완전히 같기 때문에 복소기하에서는 Wirtinger 미분을 기본으로 다룬다고 한다.
정칙함수에 대해서 f : C → C f : \mathbb{C} \to \mathbb{C} f : C → C 정칙 함수 라고 하자. 그러면 Wirtinger 도함수는 다음과 같다.
∂ f ∂ z = 1 2 ( ∂ ∂ x − i ∂ ∂ y ) ( u + i v ) = 1 2 ( u x + v y + i ( − u y + v x ) )
\begin{align*}
\dfrac{\partial f}{\partial z} &= \dfrac{1}{2} \left( \dfrac{\partial }{\partial x} -i \dfrac{\partial }{\partial y}\right) (u + iv)
\\ &= \dfrac{1}{2} \left( u_{x} + v_{y} + i(-u_{y} + v_{x}) \right)
\end{align*}
∂ z ∂ f = 2 1 ( ∂ x ∂ − i ∂ y ∂ ) ( u + i v ) = 2 1 ( u x + v y + i ( − u y + v x ) )
미분가능한 복소함수는 코시-리만 방정식 을 만족하므로 위 식은 다음과 같다.
∂ f ∂ z = 1 2 ( u x + u x + i ( v x + v x ) ) = u x + i v x
\dfrac{\partial f}{\partial z} = \dfrac{1}{2} \left( u_{x} + u_{x} + i(v_{x} + v_{x}) \right) = u_{x} + i v_{x}
∂ z ∂ f = 2 1 ( u x + u x + i ( v x + v x ) ) = u x + i v x
그런데 복소함수의 도함수는 f ′ = d f d z = u x + i v x f^{\prime} = \dfrac{df}{dz} = u_{x} + iv_{x} f ′ = d z df = u x + i v x
∂ f ∂ z = u x + i v x = d f d z
\dfrac{\partial f}{\partial z} = u_{x} + i v_{x} = \dfrac{df}{dz}
∂ z ∂ f = u x + i v x = d z df
따라서 미분가능한 함수 f f f ∂ ∂ z \dfrac{\partial }{\partial z} ∂ z ∂ d d z \dfrac{d}{dz} d z d ∂ f ∂ z ˉ = 0 \dfrac{\partial f}{\partial \bar{z}}=0 ∂ z ˉ ∂ f = 0
∂ f ∂ z ˉ = 1 2 ( ∂ ∂ x + i ∂ ∂ y ) ( u + i v ) = 1 2 ( u x − v y + i ( u y + v x ) ) = 0
\begin{align*}
\dfrac{\partial f}{\partial \bar{z}} &= \dfrac{1}{2} \left( \dfrac{\partial }{\partial x} +i \dfrac{\partial }{\partial y}\right) (u + iv)
\\ &= \dfrac{1}{2} \left( u_{x} - v_{y} + i(u_{y} + v_{x}) \right)
\\ &= 0
\end{align*}
∂ z ˉ ∂ f = 2 1 ( ∂ x ∂ + i ∂ y ∂ ) ( u + i v ) = 2 1 ( u x − v y + i ( u y + v x ) ) = 0
실수부와 허수부가 모두 0 0 0
⟹ { u x = v y u y = − v x
\implies \begin{cases}
u_{x} = v_{y}
\\ u_{y} = - v_{x}
\end{cases}
⟹ { u x = v y u y = − v x
이는 코시-리만 방정식 과 같다. 따라서 ∂ f ∂ z ˉ = 0 \dfrac{\partial f}{\partial \bar{z}}=0 ∂ z ˉ ∂ f = 0 f f f
f f f ∂ f ∂ z ‾ = 0 \dfrac{\partial f}{\partial \overline{z}} = 0 ∂ z ∂ f = 0 f f f z ‾ \overline{z} z 비정칙함수에 대해서 z ‾ \overline{z} z f ( z ) = ∣ z ∣ f(z) = \left| z \right| f ( z ) = ∣ z ∣ 최적화 하기위해 변화율을 알고싶다고 하자. 하지만 f f f d f d z \dfrac{df}{dz} d z df f f f
∂ f ∂ z = ∂ z z ‾ ∂ z = z ‾
\dfrac{\partial f}{\partial z} = \dfrac{\partial z\overline{z}}{\partial z} = \overline{z}
∂ z ∂ f = ∂ z ∂ z z = z
실제로 통신 등의 공학에서 이러한 테크닉을 사용한다.
성질 미분이라고 불리기 위해서 응당히 가져야할 성질들도 잘 갖고 있음을 확인할 수 있다.
선형성:
∂ ( a f + g ) ∂ z = a ∂ f ∂ z + ∂ g ∂ z
\dfrac{\partial (af + g)}{\partial z} = a\dfrac{\partial f}{\partial z} + \dfrac{\partial g}{\partial z}
∂ z ∂ ( a f + g ) = a ∂ z ∂ f + ∂ z ∂ g
∂ ( a f + g ) ∂ z ‾ = a ∂ f ∂ z ‾ + ∂ g ∂ z ‾
\dfrac{\partial (af + g)}{\partial \overline{z}} = a\dfrac{\partial f}{\partial \overline{z}} + \dfrac{\partial g}{\partial \overline{z}}
∂ z ∂ ( a f + g ) = a ∂ z ∂ f + ∂ z ∂ g
곱의 미분:
∂ ( f g ) ∂ z = ∂ f ∂ z g + f ∂ g ∂ z
\dfrac{\partial (fg)}{\partial z} = \dfrac{\partial f}{\partial z}g + f\dfrac{\partial g}{\partial z}
∂ z ∂ ( f g ) = ∂ z ∂ f g + f ∂ z ∂ g
∂ ( f g ) ∂ z ‾ = ∂ f ∂ z ‾ g + f ∂ g ∂ z ‾
\dfrac{\partial (fg)}{\partial \overline{z}} = \dfrac{\partial f}{\partial \overline{z}}g + f\dfrac{\partial g}{\partial \overline{z}}
∂ z ∂ ( f g ) = ∂ z ∂ f g + f ∂ z ∂ g
연쇄 법칙
∂ ( f ∘ g ) ∂ z = ∂ f ∂ w ∂ g ∂ z + ∂ f ∂ w ‾ ∂ g ‾ ∂ z
\dfrac{\partial (f \circ g)}{\partial z} = \dfrac{\partial f}{\partial w} \dfrac{\partial g}{\partial z} + \dfrac{\partial f}{\partial \overline{w}} \dfrac{\partial \overline{g}}{\partial z}
∂ z ∂ ( f ∘ g ) = ∂ w ∂ f ∂ z ∂ g + ∂ w ∂ f ∂ z ∂ g
∂ ( f ∘ g ) ∂ z ‾ = ∂ f ∂ w ∂ g ∂ z ‾ + ∂ f ∂ w ‾ ∂ g ‾ ∂ z ‾
\dfrac{\partial (f \circ g)}{\partial \overline{z}} = \dfrac{\partial f}{\partial w} \dfrac{\partial g}{\partial \overline{z}} + \dfrac{\partial f}{\partial \overline{w}} \dfrac{\partial \overline{g}}{\partial \overline{z}}
∂ z ∂ ( f ∘ g ) = ∂ w ∂ f ∂ z ∂ g + ∂ w ∂ f ∂ z ∂ g
이것 외에도,
∂ 2 f ∂ z ∂ z ‾ = 1 4 ( ∂ 2 f ∂ x 2 + ∂ 2 f ∂ y 2 ) = 1 4 Δ f
\dfrac{\partial ^{2} f}{\partial z \partial \overline{z}} = \dfrac{1}{4} \left( \dfrac{\partial ^{2} f}{\partial x^{2}} + \dfrac{\partial ^{2} f}{\partial y^{2}}\right) = \dfrac{1}{4}\Delta f
∂ z ∂ z ∂ 2 f = 4 1 ( ∂ x 2 ∂ 2 f + ∂ y 2 ∂ 2 f ) = 4 1 Δ f
d ( f ∘ ϕ ) d t = ∂ f ∂ z ∂ ϕ ∂ t + ∂ f ∂ z ‾ ∂ ϕ ‾ ∂ t
\dfrac{d (f \circ \phi)}{d t} = \dfrac{\partial f}{\partial z} \dfrac{\partial \phi}{\partial t} + \dfrac{\partial f}{\partial \overline{z}} \dfrac{\partial \overline{\phi}}{\partial t}
d t d ( f ∘ ϕ ) = ∂ z ∂ f ∂ t ∂ ϕ + ∂ z ∂ f ∂ t ∂ ϕ
이때 ϕ : R → C \phi : \mathbb{R} \to \mathbb{C} ϕ : R → C
