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2008-05-18T19:17:38Z

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Vipul: New page: ==Statement== ===Complex-differentiable at a point=== Suppose $U \subset \mathbb{C}$ is an open subset and $f,g:U \to \mathbb{C}$ are functions. Suppose $z_0 \...$

2008-04-26T21:02:29Z

New page: ==Statement== ===Complex-differentiable at a point=== Suppose <math>U \subset \mathbb{C}</math> is an open subset and <math>f,g:U \to \mathbb{C}</math> are functions. Suppose <math>z_0 \...

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==Statement==

===Complex-differentiable at a point===

Suppose <math>U \subset \mathbb{C}</math> is an open subset and <math>f,g:U \to \mathbb{C}</math> are functions. Suppose <math>z_0 \in U</math> is a point such that <math>f,g</math> are both [[function complex-differentiable at a point|complex-differentiable]] at <math>z_0</math>. Define <math>fg:U \to \mathbb{C}</math> as:

<math>fg := z \mapsto f(z)g(z)</math>

Then <math>fg</math> is complex-differentiable at <math>z_0</math> and:

<math>(fg)'(z_0) = f'(z_0)g(z_0) + f(z_0)g'(z_0)</math>

===For holomorphic functions===

Suppose <math>U \subset \mathbb{C}</math> is an open subset and <math>f,g:U \to \mathbb{C}</math> are [[holomorphic function]]s. Then the function <math>fg: U \to \mathbb{C}</math> given by:

<math>fg := z \mapsto f(z)g(z)</math>

is also a holomorphic function, and for any <math>z \in U</math>, we have:

<math>(fg)'(z) = f'(z)g(z) + f(z)g'(z)</math>

Product rule for complex differentiation - Revision history

Vipul: 1 revision

Vipul: New page: ==Statement== ===Complex-differentiable at a point=== Suppose U \subset \mathbb{C} is an open subset and f,g:U \to \mathbb{C} are functions. Suppose z_0 \...

Vipul: New page: ==Statement== ===Complex-differentiable at a point=== Suppose $U \subset \mathbb{C}$ is an open subset and $f,g:U \to \mathbb{C}$ are functions. Suppose $z_0 \...$