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Math 411: Honours Complex Variables - University of Alberta

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10 CHAPTER 2. COMPLEX DIFFERENTIATION<br />

Pro<strong>of</strong>. As over R.<br />

Proposition 2.3. Let D,E ⊂ C, let g : D → C and f : E → C be such that<br />

g(D) ⊂ E, and let z0 ∈ intD be such that w0 := g(z0) ∈ intE. Further, suppose<br />

that g is complex differentiable at z0 and f is complex differentiable at w0. Then f◦g<br />

is complex differentiable at z0 with<br />

Pro<strong>of</strong>. As over R.<br />

Examples.<br />

(f ◦g) ′ (z0) = f ′ (g(z0))g ′ (z0).<br />

1. All constant functions are (on all <strong>of</strong> C) complex differentiable, as is z ↦→ z on<br />

C. Consequently, all complex polynomials are complex differentiable on all <strong>of</strong><br />

C, and rational functions are complex differentiable wherever they are defined.<br />

2. Let<br />

f: C → C, z ↦→ ¯z,<br />

and let z0 = x0+iy0 ∈ C. Assume that f is complex differentiable at z0. Then<br />

we have<br />

as well as<br />

f ′ ¯z − ¯z0<br />

(z0) = lim<br />

z→z0<br />

z −z0<br />

(x0 −iy)−(x0 −iy0)<br />

= lim<br />

y→y0 (x0 +iy)−(x0 +iy0)<br />

i(y0 −y)<br />

= lim<br />

y→y0 i(y −y0)<br />

= −1<br />

f ′ ¯z − ¯z0<br />

(z0) = lim<br />

z→z0<br />

z −z0<br />

(x−iy0)−(x0 −iy0)<br />

= lim<br />

x→x0 (x+iy0)−(x0 +iy0)<br />

x−x0<br />

= lim<br />

x→x0 x−x0<br />

= 1,<br />

which is impossible. Hence, f is not complex differentiable at any z0 ∈ C. (On<br />

the other hand, f is continuously partially differentiable—as a function <strong>of</strong> two<br />

real variables—on all <strong>of</strong> C.)<br />

Lemma 2.1. The following are equivalent for an R-linear map T : C → C:

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