Math 411: Honours Complex Variables - University of Alberta
Math 411: Honours Complex Variables - University of Alberta
Math 411: Honours Complex Variables - University of Alberta
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12 CHAPTER 2. COMPLEX DIFFERENTIATION<br />
(ii) f is totally differentiable at z0 (in the sense <strong>of</strong> multivariable calculus), and the<br />
Cauchy–Riemann differential equations<br />
hold.<br />
Pro<strong>of</strong>. (i) =⇒ (ii): Define<br />
∂u<br />
∂x (z0) = ∂v<br />
∂y (z0) and<br />
T: C → C, z ↦→ f ′ (z0)z,<br />
∂u<br />
∂y (z0) = − ∂v<br />
∂x (z0)<br />
and note that<br />
�<br />
|f(z)−f(z0)−T(z −z0)| �<br />
= �<br />
f(z)−f(z0)<br />
|z −z0| � −f<br />
z −z0<br />
′ �<br />
�<br />
(z0) �<br />
� → 0<br />
as z → z0. Therefore, f is totally differentiable at z0. From multivariable calculus, it<br />
follows that the matrix representation <strong>of</strong> T with respect to the standard basis <strong>of</strong> R is<br />
the Jacobian <strong>of</strong> f, i.e. � �<br />
ux(z0) uy(z0)<br />
Jf(z0) = .<br />
vx(z0) vy(z0)<br />
Since T is C-linear, Lemma 2.1 yields that<br />
ux(z0) = vy(z0) and uy(z0) = −vx(z0).<br />
(ii) =⇒ (i): Since f is totally differentiable at z0, we have a unique R-linear map<br />
T : C → C such that<br />
|f(z)−f(z0)−T(z −z0)|<br />
lim<br />
= 0.<br />
z→z0 |z −z0|<br />
Asweknowfrommultivariablecalculus, T isrepresented byJf(z0)withrespect tothe<br />
standard basis <strong>of</strong> R2 . Since the Cauchy–Riemann differential equations are supposed<br />
to hold, Jf(z0) is <strong>of</strong> the form described in Lemma 2.1(iv). By Lemma 2.1, there thus<br />
exists c ∈ C such that T(z) = cz for all z ∈ C. It follows that<br />
� �<br />
�<br />
�<br />
f(z)−f(z0) �<br />
� −c�<br />
z � −z0<br />
= |f(z)−f(z0)−c(z −z0)|<br />
→ 0<br />
|z −z0|<br />
as z → z0. Hence, f is complex differentiable at z0.<br />
Remark. In the situation <strong>of</strong> Theorem 2.1, we have<br />
f ′ � �� �<br />
ux(z0) uy(z0) 1<br />
(z0)1 =<br />
= ux(z0)+ivx(z0)<br />
vx(z0) vy(z0) 0<br />
as well as<br />
so that<br />
f ′ (z0)i =<br />
� ux(z0) uy(z0)<br />
vx(z0) vy(z0)<br />
�� 0<br />
1<br />
�<br />
= uy(z0)+ivy(z0),<br />
f ′ (z0) = ux(z0)+ivx(z0) = vy(z0)−iuy(z0).