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Holomorphe Ableitung/z und z konjugiert/Beispiel
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Es ist
∂
z
∂
z
=
∂
(
x
+
i
y
)
∂
z
=
1
2
⋅
∂
(
x
+
i
y
)
∂
x
−
i
2
⋅
∂
(
x
+
i
y
)
∂
y
=
1
2
−
i
2
i
=
1
{\displaystyle {}{\frac {\partial z}{\partial z}}={\frac {\partial (x+{\mathrm {i} }y)}{\partial z}}={\frac {1}{2}}\cdot {\frac {\partial (x+{\mathrm {i} }y)}{\partial x}}-{\frac {\mathrm {i} }{2}}\cdot {\frac {\partial (x+{\mathrm {i} }y)}{\partial y}}={\frac {1}{2}}-{\frac {\mathrm {i} }{2}}{\mathrm {i} }=1\,}
und
∂
z
¯
∂
z
=
∂
(
x
−
i
y
)
∂
z
=
1
2
⋅
∂
(
x
−
i
y
)
∂
x
−
i
2
⋅
∂
(
x
−
i
y
)
∂
y
=
1
2
+
i
2
i
=
0
.
{\displaystyle {}{\frac {\partial {\overline {z}}}{\partial z}}={\frac {\partial (x-{\mathrm {i} }y)}{\partial z}}={\frac {1}{2}}\cdot {\frac {\partial (x-{\mathrm {i} }y)}{\partial x}}-{\frac {\mathrm {i} }{2}}\cdot {\frac {\partial (x-{\mathrm {i} }y)}{\partial y}}={\frac {1}{2}}+{\frac {\mathrm {i} }{2}}{\mathrm {i} }=0\,.}
