ウィルティンガーの微分
補題1. f , g ∈ C 1 ( Ω ) {\displaystyle f,g\in C^{1}(\Omega )} とし、 α , β {\displaystyle \alpha ,\beta } を複素数とすると、 i = 1 , ... , n {\displaystyle i=1,\dots ,n} に対して、以下の等式が成り立つ
∂ ∂ z i ( α f + β g ) = α ∂ f ∂ z i + β ∂ g ∂ z i , ∂ ∂ z ̄ i ( α f + β g ) = α ∂ f ∂ z ̄ i + β ∂ g ∂ z ̄ i {\displaystyle {\frac {\partial }{\partial z_{i}}}\left(\alpha f+\beta g\right)=\alpha {\frac {\partial f}{\partial z_{i}}}+\beta {\frac {\partial g}{\partial z_{i}}},\quad {\frac {\partial }{\partial {\bar {z}}_{i}}}\left(\alpha f+\beta g\right)=\alpha {\frac {\partial f}{\partial {\bar {z}}_{i}}}+\beta {\frac {\partial g}{\partial {\bar {z}}_{i}}}}
補題4. f ∈ C 1 ( Ω ) {\displaystyle f\in C^{1}(\Omega )} であれば、 i = 1 , ... , n {\displaystyle i=1,\dots ,n} に対して、以下の等式が成り立つ
∂ f ̄ ∂ z i = ∂ f ̄ ∂ z ̄ i , ∂ f ̄ ∂ z ̄ i = ∂ f ̄ ∂ z i {\displaystyle {\frac {\overline {\partial f}}{\partial z_{i}}}={\frac {\partial {\bar {f}}}{\partial {\bar {z}}_{i}}},\quad {\frac {\overline {\partial f}}{\partial {\bar {z}}_{i}}}={\frac {\partial {\bar {f}}}{\partial z_{i}}}}