Holomorphic and Conformal Maps

Holomorphic Functions

Domain

A domain is a connected open subset of .

Holomorphic Function

A function is called holomorphic on domain if it is -differentiable at every point in . We write for the set of functions that are holomorphic on . We call a function an entire function.

e.g. The function is entire. Polynomials and , are also entire functions.

Proposition

yields a ring with pointwise addition and multiplication.

Proof For all , and are also holomorphic. The additive identity is the zero function. The associative and distributive properties can be checked directly from the definition.

Biholomorphism and Conformally Equivalence

Let be domains. A function is called biholomorphic (or a conformal equivalence) if is a bijection, and both and are holomorphic. Two domains are called conformally equivalent if there exists a conformal equivalence between them.

Proposition

Conformal equivalence is an equivalence relation on the set of domains.

Proposition

If is a biholomorphism, the map given by is an isomorphism of rings.

Proof To see that it is an isomorphism, we can check that . For any arbitrary , we have Thus is the inverse of .

Holomorphic Branch of The Logarithm

If is a domain, a holomorphic branch of the logarithm on is a function such that for all .

e.g. is not a holomorphic branch of the logarithm on , as it is not continuous thus not holomorphic.

To find holomorphic branches of the logarithm, we think about which domains the function is injective on. For , define the strip . Recall the proposition that iff for , so when , is injective. The image of is the wedge bounded by the angles and , which we write as . By the theorem above, such is biholomorphic, and its inverse exists. We denote this inverse as , then we have Proof Since we know that is the inverse of , we have . So since the inverse is unique if it exists. In fact are holomorphic branches of the logarithm on .

Conformal Maps

Lemma

Let , then the following are equivalent:

• for some not both zero .
• is the product of a rotation and a scaling for .
• For any non-zero , the angle from to is the same as the angle from to .

Proof We first show 1 implies 2. If is of the given form, then with . To see 2 implies 1, we have 2 implies 3 is immediate since both scaling and rotations preserve angles. Now we show 3 implies 2. Suppose preserves angles. Let be the first standard basis vector in . Let be an angle corresponding to the non-zero vector . Then we have for some . Since preserves angles, also preserves angles. If is any non-zero vector in , the angle from to is the same as the angle from to , which means that is on the same ray as , i.e. is an eigenvector of . As the above holds for any non-zero vector , is a scaling matrix , so .

Comment

So the matrices corresponding to the Cauchy-Riemann equations are exactly the angle-preserving linear transformations. Now it is necessary to talk about angle-preserving maps that are not linear.

Path and Angle

A path in a domain is a continuous function . If and are two paths in and , then we say that and cross at . If we have such crossing paths are non-zero, we define the angle from to at as the angle from tangent vectors to in .

Conformal Map

Let be an differentiable function, and let . Then is conformal at if whenever and are two paths in that cross at , the angle from to at is the same as the angle from to at . A function is conformal if it is conformal at every point in .

Theorem

Let be an -differentiable function, and let . Then is conformal at if and only if is -differentiable at and . So is conformal if and only if is holomorphic and at every point .

Proof Let and be two crossing paths at . Then the angle from to at is the angle from to in . By the chain rule, we have and similarly for . So the angle is preserved if and only if is of the form for some not both zero , which is equivalent to is -differentiable and .

e.g. The function is conformal on . Instead, reverses angles.

Corollary

Biholomorphic maps are conformal.

Proof Suppose is a biholomorphism. Then is a bijection, and both and are holomorphic, we have Now differentiate both sides, we obtain Thus , hence is conformal.