Komplexe Zahlen/Reell-differenzierbare Funktionen/Holomorphe Ableitung/Quotientenregel/Fakt/Beweis

Beweis

Wir verwenden die Quotientenregel für die beiden partielle Ableitungen, also

${}{\frac {\partial \left({\frac {f}{g}}\right)}{\partial x}}={\frac {g{\frac {\partial f}{\partial x}}-f{\frac {\partial g}{\partial x}}}{g^{2}}}\,$

und

${}{\frac {\partial \left({\frac {f}{g}}\right)}{\partial y}}={\frac {g{\frac {\partial f}{\partial y}}-f{\frac {\partial g}{\partial y}}}{g^{2}}}\,,$

die in dieser Form auf Aufgabe beruht. Somit ist

${}{\begin{aligned}{\frac {\partial \left({\frac {f}{g}}\right)}{\partial z}}&={\frac {1}{2}}\cdot {\frac {\partial \left({\frac {f}{g}}\right)}{\partial x}}-{\frac {\mathrm {i} }{2}}\cdot {\frac {\partial \left({\frac {f}{g}}\right)}{\partial y}}\\&={\frac {1}{2}}\cdot {\frac {g{\frac {\partial f}{\partial x}}-f{\frac {\partial g}{\partial x}}}{g^{2}}}-{\frac {\mathrm {i} }{2}}\cdot {\frac {g{\frac {\partial f}{\partial y}}-f{\frac {\partial g}{\partial y}}}{g^{2}}}\\&={\frac {1}{g^{2}}}{\left(g{\left({\frac {1}{2}}\cdot {\frac {\partial f}{\partial x}}-{\frac {\mathrm {i} }{2}}\cdot {\frac {\partial f}{\partial y}}\right)}-f{\left({\frac {1}{2}}\cdot {\frac {\partial g}{\partial x}}-{\frac {\mathrm {i} }{2}}\cdot {\frac {\partial g}{\partial y}}\right)}\right)}\\&={\frac {1}{g^{2}}}{\left(g{\frac {\partial f}{\partial z}}-f{\frac {\partial g}{\partial z}}\right)}.\end{aligned}}$

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