Usuari:Laplaci/proves

Diferenciabilitat de funcions de variables complexes i equacions de Cauchy-Riemann

Donada un variable complexa

$z=x+iy$

I la funció $f(z):\mathbb {C} \rightarrow \mathbb {C}$

$f(z)=u(z)+iv(z)$

Llavors la seva derivada total serà de la forma

${\operatorname {d} \!f \over \operatorname {d} \!z}={\partial f \over \partial u}{\partial u \over \partial z}+{\partial f \over \partial v}{\partial v \over \partial z}={\partial u \over \partial z}+{i{\partial v \over \partial z}}$

Com que $z$ té una component real i una imaginària, les analitzem per separat

• ${\text{ Eix real: }}z=x\rightarrow \partial z=\partial x$
• ${\text{ Eix imaginari: }}z=iy\rightarrow \partial z=i\partial y$

Per tant, recordant que ${1/i}={-i}$

• ${\text{ Eix real: }}{\operatorname {d} \!f \over \operatorname {d} \!z}={\partial u \over \partial z}+{i{\partial v \over \partial z}}={\partial u \over \partial x}+i{\partial v \over \partial x}$
• ${\text{ Eix imaginari: }}{\operatorname {d} \!f \over \operatorname {d} \!z}={\partial u \over \partial z}+{i{\partial v \over \partial z}}=-i{\partial u \over \partial y}+{\partial v \over \partial y}$

Perquè la derivada existeixi, les seves derivades en les direccions de l'eix real i imaginari han d'existir i ser iguals

${{\partial u \over \partial x}+i{\partial v \over \partial x}}={\partial v \over \partial y}-{i{\partial u \over \partial y}}$

Tenint en compte que

${\partial v \over \partial y}-{i{\partial u \over \partial y}}={{1 \over i}\cdot i}{\Bigl (}{\partial v \over \partial y}-{i{\partial u \over \partial y}}{\Bigr )}={1 \over i}{\Bigl (}{\partial u \over \partial y}+{i{\partial v \over \partial y}}{\Bigr )}$

Podem expressar aquesta igualtat respecte a $f$ de la forma

${{\partial u \over \partial x}+i{\partial v \over \partial x}}={1 \over i}{\Bigl (}{\partial u \over \partial y}+{i{\partial v \over \partial y}}{\Bigr )}$

$i{\partial f \over \partial x}={\partial f \over \partial y}$

o igualant les components reals i imaginàries

${\partial u \over \partial x}={\partial v \over \partial y}$

${\partial v \over \partial x}=-{\partial u \over \partial y}$