1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

104 CHAPTER 3 • ANALYTIC AND HARMONIC FUNC'tlONS

• EXAMPLE 3.6 Show that the function defined by

is not differentiable at the point z 0 = 0 even though the Cauchy-Riemann equa-

tions are satisfied at (0, 0).

Solution We must use limits to calculate the partial derivatives at (0, 0).

x^3 - 0

(0 0 )

_

1

. u (x, 0) - u (0, 0)
1
. x•+o
Ux , - 1m Q - Im -^1.
x-o x - x-o x

Similarly, we can show that

Uy (0, 0) = 0, 11., (0, O) = 0 and 1111 (0,0) = 1.

Hence the Cauchy-Riemann equations hold at the point (O,O).

We now show that f is not differentiable at zo = 0. Letting z approach 0

along the x-axis gives

lim f (x + Oi) -f (0) = lim "'.,

2

• 0 = lim x - 0 = l.

(x,o)-(0.0) x + Oi - 0 ,,_o x - 0 >:-0 x - 0

But if we let z approach 0 along the line y = x given by the parametric equations

x = t and y = t , then

• 2t^3. (-21^3 )
  . f(t+it)-f(O). ""2t'+ i ur
  hm.
  0

= hm.

{t,t )- (O,o) t +it - t-o t +it

• t -it
  =Jim---= - 1.
  t-0 t +it

The two limits are distinct, so f is not differentiable at the origin.

Example 3.6 reiterates that the mere satisfaction of the Cauchy-Riemann

equations is not sufficient to guarantee the differentiability of a function. The

following theorem, however, gives conditions that guarantee the differentiability

off at zo, so that we can use Equation (3-14) or (3-15) to compute f' (zo). They

are referred to as the Cauchy- Riemann conditions for differentiability.