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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).

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