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Complex Analysis - Maths KU

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156 Chapter 3 Analytic Functions<br />

EXAMPLE 4 A Function Differentiable on a Line<br />

In Example 2 we saw that the complex function f(z) =2x 2 + y + i(y 2 − x)<br />

was nowhere analytic, but yet the Cauchy-Riemann equations were satisfied<br />

on the line y =2x. But since the functions u(x, y) =2x 2 + y, ∂u/∂x =4x,<br />

∂u/∂y =1,v(x, y) =y 2 − x, ∂v/∂x = −1 and ∂v/∂y =2y are continuous at<br />

every point, it follows that f is differentiable on the line y =2x. Moreover,<br />

from (9) we see that the derivative of f at points on this line is given by<br />

f ′ (z) =4x − i =2y − i.<br />

The following theorem is a direct consequence of the Cauchy-Riemann<br />

equations. Its proof is left as an exercise. See Problems 29 and 30 in Exercises<br />

3.2.<br />

Theorem 3.6 Constant Functions<br />

Suppose the function f(z) =u(x, y)+iv(x, y) is analytic in a domain D.<br />

(i) If|f (z)| is constant in D, then so is f(z).<br />

(ii) Iff ′ (z) =0inD, then f(z) =c in D, where c is a constant.<br />

Polar Coordinates In Section 2.1 we saw that a complex function<br />

can be expressed in terms of polar coordinates. Indeed, the form f(z) =<br />

u(r, θ) +iv(r, θ) is often more convenient to use. In polar coordinates the<br />

Cauchy-Riemann equations become<br />

∂u<br />

∂r<br />

1 ∂v<br />

=<br />

r ∂θ ,<br />

∂v<br />

∂r<br />

= −1<br />

r<br />

∂u<br />

. (10)<br />

∂θ<br />

The polar version of (9) at a point z whose polar coordinates are (r, θ) is<br />

then<br />

f ′ (z) =e −iθ<br />

� �<br />

∂u<br />

+ i∂v =<br />

∂r ∂r<br />

1<br />

r e−iθ<br />

� �<br />

∂v<br />

− i∂u . (11)<br />

∂θ ∂θ<br />

See Problems 33 and 34 in Exercises 3.2.<br />

Remarks Comparison with Real <strong>Analysis</strong><br />

In real calculus, one of the noteworthy properties of the exponential function<br />

f(x) =e x is that f ′ (x) =e x . In (3) of Section 2.1 we gave the definition<br />

of the complex exponential f(z) =e z . We are now in a position<br />

to show that f(z) =e z is differentiable everywhere and that this complex<br />

function shares the same derivative property as its real counterpart, that<br />

is, f ′ (z) =f(z). See Problem 25 in Exercises 3.2.

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