¿por qué$\frac{df}{dz}=\frac{1}{2}\left ( \frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y}\right )$, ¿por qué la $\frac{1}{2}$?

Question

Tengo problemas para ver donde el $\frac{1}{2}$ proviene de la en $$\frac{df}{dz}=\frac{1}{2}\left ( \frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y}\right )$$

Para un cambio de variables $z=x+iy$ hemos $$\frac{df}{dz}= \ \frac{\partial f}{\partial x}\frac{\partial x}{\partial z}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial z}$$ and $\frac{\partial x}{\partial z}=1$ and $\frac{\partial y}{\partial z}=-i$. Therefore we have the above but without the $\frac{1}{2}$. I've seen someone derive the correct expression by including the change of variables for $\overline{z}$ sin embargo no veo la manera de que sea necesario, se debe trabajar sin, derecho? No sé de qué me estoy perdiendo?

Answer 1

Así se inicia con la fórmula general para el diferencial $$df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy.$$ In complex analysis we prefer to use $dz$ and $d\overline{z}$. So using, $dx = \dfrac{dz+d\overline{z}}{2}$ and $dy = \dfrac{dz-d\overline{z}}{2i},$ one finally gets $$df = \frac{1}{2}\left ( \frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y}\right ) dz + \frac{1}{2}\left ( \frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}\right )d\overline{z}.$$ Now it is natural to define $\frac{\partial}{\partial z}$ to be $\frac{1}{2}\left ( \frac{\partial }{\partial x}-i\frac{\partial }{\partial y}\right )$ and $\frac{\partial}{\partial \overline{z}}$ to be $\frac{1}{2}\left ( \frac{\partial }{\partial x}+i\frac{\partial }{\partial y}\right )$. Thanks to that we have the beautiful formula $$df = \frac{\partial f}{\partial z} dz + \frac{\partial f}{\partial \overline{z}} d\overline{z}.$$

Answer 2

Es solo una convención, no hay una profunda razón matemática en el factor de $\frac{1}{2}$.

Creo que el convenio viene de $dz=dx+idy$(Generalmente preferimos $n$-formularios en lugar de $n$-tensor de vectores tangente, ¿no?) y luego tomar la base dual.