Diferenciabilitat de funcions de variables complexes i equacions de Cauchy-Riemann
modifica
Donada un variable complexa
z
=
x
+
i
y
{\displaystyle z=x+iy}
I la funció
f
(
z
)
:
C
→
C
{\displaystyle f(z):\mathbb {C} \rightarrow \mathbb {C} }
f
(
z
)
=
u
(
z
)
+
i
v
(
z
)
{\displaystyle f(z)=u(z)+iv(z)}
Llavors la seva derivada total serà de la forma
d
f
d
z
=
∂
f
∂
u
∂
u
∂
z
+
∂
f
∂
v
∂
v
∂
z
=
∂
u
∂
z
+
i
∂
v
∂
z
{\displaystyle {\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
{\displaystyle z}
Eix real:
z
=
x
→
∂
z
=
∂
x
{\displaystyle {\text{ Eix real: }}z=x\rightarrow \partial z=\partial x}
Eix imaginari:
z
=
i
y
→
∂
z
=
i
∂
y
{\displaystyle {\text{ Eix imaginari: }}z=iy\rightarrow \partial z=i\partial y}
Per tant, recordant que
1
/
i
=
−
i
{\displaystyle {1/i}={-i}}
Eix real:
d
f
d
z
=
∂
u
∂
z
+
i
∂
v
∂
z
=
∂
u
∂
x
+
i
∂
v
∂
x
{\displaystyle {\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}}
Eix imaginari:
d
f
d
z
=
∂
u
∂
z
+
i
∂
v
∂
z
=
−
i
∂
u
∂
y
+
∂
v
∂
y
{\displaystyle {\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
∂
u
∂
x
+
i
∂
v
∂
x
=
∂
v
∂
y
−
i
∂
u
∂
y
{\displaystyle {{\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
∂
v
∂
y
−
i
∂
u
∂
y
=
1
i
⋅
i
(
∂
v
∂
y
−
i
∂
u
∂
y
)
=
1
i
(
∂
u
∂
y
+
i
∂
v
∂
y
)
{\displaystyle {\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
{\displaystyle f}
∂
u
∂
x
+
i
∂
v
∂
x
=
1
i
(
∂
u
∂
y
+
i
∂
v
∂
y
)
{\displaystyle {{\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
∂
f
∂
x
=
∂
f
∂
y
{\displaystyle i{\partial f \over \partial x}={\partial f \over \partial y}}
o igualant les components reals i imaginàries
∂
u
∂
x
=
∂
v
∂
y
{\displaystyle {\partial u \over \partial x}={\partial v \over \partial y}}
∂
v
∂
x
=
−
∂
u
∂
y
{\displaystyle {\partial v \over \partial x}=-{\partial u \over \partial y}}
