Let us now define what complex differentiability is.
- Example 2.3.2
The function
![{\displaystyle f:\mathbb {C} \to \mathbb {C} ,f(z)={\bar {z}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/344d98b48904acf596e0f0665c135f46f2ce3684)
is nowhere complex differentiable.
- Proof
Let
be arbitrary. Assume that
is complex differentiable at
, i.e. that
![{\displaystyle \lim _{z\to z_{0} \atop z\in \mathbb {C} }{\frac {{\bar {z}}-{\bar {z}}_{0}}{z-z_{0}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/306c802672489718544a424b08700c0759fddd36)
exists.
We choose
![{\displaystyle {\begin{aligned}A:=\{z\in \mathbb {C} |\Re z=\Re z_{0}\}\\B:=\{z\in \mathbb {C} |\Im z=\Im z_{0}\}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/56c463887ba975f4f659b53a5306889f12aca2c6)
Due to lemma 2.2.3, which is applicable since of course
is open, we have:
![{\displaystyle \lim _{z\to z_{0} \atop z\in A}{\frac {{\bar {z}}-{\bar {z}}_{0}}{z-z_{0}}}=\lim _{z\to z_{0} \atop z\in \mathbb {C} }{\frac {{\bar {z}}-{\bar {z}}_{0}}{z-z_{0}}}=\lim _{z\to z_{0} \atop z\in B}{\frac {{\bar {z}}-{\bar {z}}_{0}}{z-z_{0}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0155b4c7dc87cee3d65d0d6f73c24d4c6b1d50ee)
But
![{\displaystyle {\begin{aligned}&\lim _{z\to z_{0} \atop z\in A}{\frac {{\bar {z}}-{\bar {z}}_{0}}{z-z_{0}}}=\lim _{z\to z_{0} \atop z\in A}{\frac {\Re (z-z_{0})-i\Im (z-z_{0})}{\Re (z-z_{0})+i\Im (z-z_{0})}}=-1\\\\&\lim _{z\to z_{0} \atop z\in B}{\frac {{\bar {z}}-{\bar {z}}_{0}}{z-z_{0}}}=\lim _{z\to z_{0} \atop z\in B}{\frac {\Re (z-z_{0})-i\Im (z-z_{0})}{\Re (z-z_{0})+i\Im (z-z_{0})}}=1\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/42f6e9d237efc315b8cfc8c60a65c68209d8a052)
a contradiction.
We can define a natural bijective function from
to
as follows:
![{\displaystyle \Phi (x+yi):=(x,y)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/993ae27997d6530fdc65a662ab8c205f5ba36baf)
In fact,
is a vector space isomorphism between
and
.
The inverse of
is given by
![{\displaystyle \Phi ^{-1}:\mathbb {R} ^{2}\to \mathbb {C} ,\Phi ^{-1}(x,y)=x+yi}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7ad232182ba516820b5c02a84ef22d53647efa55)
Theorem and definitions 2.3.3:
Let
be open, let
be a function and let
. If
is complex differentiable at
, then the functions
![{\displaystyle {\begin{aligned}&u:\Phi (O)\to \mathbb {R} ,u(x,y)=\Re f(x+yi)\\&v:\Phi (O)\to \mathbb {R} ,v(x,y)=\Im f(x+yi)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0c7f5297897fa7db5f02fe5a4a65d912e0fa7240)
are well-defined, differentiable at
and satisfy the equations
![{\displaystyle {\begin{aligned}&\partial _{x}u(x_{0},y_{0})=\partial _{y}v(x_{0},y_{0})\\&\partial _{y}u(x_{0},y_{0})=-\partial _{x}v(x_{0},y_{0})\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1257ca8d1ec556a888c1f0bd5f75c9c505a2a8ff)
These equations are called the Cauchy-Riemann equations.
- Proof
1. We prove well-definedness of
.
Let
. We apply the inverse function on both sides to obtain:
![{\displaystyle x+yi\in \Phi ^{-1}(\Phi (O))=O}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2078aa31274a41b362a1052ec1f5a9a076af0d43)
where the last equality holds since
is bijective (for any bijective
we have
if
; see exercise 1).
3. We prove differentiability of
and
and the Cauchy-Riemann equations.
We define
![{\displaystyle {\begin{aligned}S_{1}:=\{z\in \mathbb {C} :\Re (z)=\Re (z_{0})\}\cap O\\S_{2}:=\{z\in \mathbb {C} :\Im (z)=\Im (z_{0})\}\cap O\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3077c39bac9e9b6123db1b3e9b1fd1cdf0789486)
Then we have:
![{\displaystyle {\begin{aligned}\partial _{x}u(x_{0},y_{0})&=\lim _{x\to x_{0}}{\frac {u(x,y_{0})-u(x_{0},y_{0})}{x-x_{0}}}&\\&=\lim _{x\to x_{0}}{\frac {\Re {\bigl (}f(x+y_{0}i){\bigr )}-\Re {\bigl (}f(x_{0}+y_{0}i){\bigr )}}{x-x_{0}}}&\\&=\Re \left(\lim _{x\to x_{0}}{\frac {f(x+y_{0}i)-f(x_{0}+y_{0}i)}{x-x_{0}}}\right)&{\text{continuity of }}\Re \\&=\Re \left(\lim _{z\to z_{0} \atop z\in S_{2}}{\frac {f(z)-f(z_{0})}{z-z_{0}}}\right)&\\&=\Re \left(\lim _{z\to z_{0} \atop z\in S_{1}}{\frac {f(z)-f(z_{0})}{z-z_{0}}}\right)&{\text{lemma 2.2.3}}\\&=\Re \left(\lim _{y\to y_{0}}{\frac {f(x_{0}+yi)-f(x_{0}+y_{0}i)}{yi-y_{0}i}}\right)&\\&=\Re \left((-i)\lim _{y\to y_{0}}{\frac {f(x_{0}+yi)-f(x_{0}+y_{0}i)}{y-y_{0}}}\right)&i^{-1}=-i\\&=\Im \left(\lim _{y\to y_{0}}{\frac {f(x_{0}+yi)-f(x_{0}+y_{0}i)}{y-y_{0}}}\right)&\\&=\partial _{y}v(x_{0},y_{0})\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6d95e7d4760633d9186033e79fe54d0d2ff89fca)
From these equations follows the existence of
, since for example
![{\displaystyle \lim _{z\to z_{0} \atop z\in S_{2}}{\frac {f(z)-f(z_{0})}{z-z_{0}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dcc417901d8d8abfe4a86f3f3f688957f2ec3864)
exists due to lemma 2.2.3.
The proof for
![{\displaystyle \partial _{y}u(x_{0},y_{0})=-\partial _{x}v(x_{0},y_{0})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6d7a60b76d6d981c7a409ea5fe79854489dc9cb4)
and the existence of
we leave for exercise 2.
- Let
be sets such that
, and let
be a bijective function. Prove that
.
- Let
be open, let
be a function and let
. Prove that if
is complex differentiable at
, then
and
exist and satisfy the equation
.
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